Answer:
c = h/(m+w)
Step-by-step explanation:
m=(h/c)-w
Add w to each side
m+w=h/c-w+w
m+w = h/c
Multiply each side by c
c(m+w) = h/c*c
c(m+w) = h
Divide each side by (m+w)
c(m+w)/(m+w) = h/(m+w)
c = h/(m+w)
Find the relationship between zeros and x-intercepts of a2 + 5a + 6.
Answer:
The x-intercepts, or zeros, which are the values of x(or a in this problem) for which the function is 0, are x = a = -2 and x = a = -3.
Step-by-step explanation:
Suppose we have a function y = f(x). The zeros, which are the values of x for which y = 0, are also called the x-intercepts of the function.
Solving a quadratic equation:
Given a second order polynomial expressed by the following equation:
[tex]ax^{2} + bx + c, a\neq0[/tex].
This polynomial has roots [tex]x_{1}, x_{2}[/tex] such that [tex]ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})[/tex], given by the following formulas:
[tex]x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}[/tex]
[tex]x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}[/tex]
[tex]\bigtriangleup = b^{2} - 4ac[/tex]
In this question:
I will write the function as a function of x, just exchanging a for x.
[tex]f(x) = x^{2} + 5x + 6[/tex]
[tex]\bigtriangleup = 5^{2} - 4*1*6 = 1[/tex]
[tex]x_{1} = \frac{-5 + \sqrt{1}}{2*1} = -2[/tex]
[tex]x_{2} = \frac{-5 - \sqrt{1}}{2*1} = -3[/tex]
The x-intercepts, or zeros, which are the values of x(or a in this problem) for which the function is 0, are x = a = -2 and x = a = -3.
What is the sum of 8 and 10?
Answer:
18
Step-by-step explanation:
pls mark brainliest
idk if this is a joke
what can you say about the end behavior of the function f(x)=-4x^6+6x^2-52
Answer:
As x gets smaller, pointing to negative infinity, the value of f decreses, pointing to negative infinity.
As x gets increases, pointing to positve infinity, the value of f decreses, pointing to negative infinity.
Step-by-step explanation:
To find the end behaviour of a function f(x), we calculate these following limits:
[tex]\lim_{x \to +\infty} f(x)[/tex]
And
[tex]\lim_{x \to -\infty} f(x)[/tex]
In this question:
[tex]f(x) = -4x^{6} + 6x^{2} - 52[/tex]
At negative infinity:
[tex]\lim_{x \to -\infty} f(x) = \lim_{x \to -\infty} -4x^{6} + 6x^{2} - 52[/tex]
When the variable points to infinity, we only consider the term with the highest exponent. So
[tex]\lim_{x \to -\infty} -4x^{6} + 6x^{2} - 52 = \lim_{x \to -\infty} -4x^{6} = -4*(-\infty)^{6} = -(\infty) = -\infty[/tex]
So as x gets smaller, pointing to negative infinity, the value of f decreses, pointing to negative infinity.
Positive infinity:
[tex]\lim_{x \to \infty} f(x) = \lim_{x \to \infty} -4x^{6} + 6x^{2} - 52 = \lim_{x \to \infty} -4x^{6} = -4*(\infty)^{6} = -(\infty) = -\infty[/tex]
So as x gets increases, pointing to positve infinity, the value of f decreses, pointing to negative infinity.
f(x) = -4x^6 +6x^2-52
The leading coefficient is negative so the left end of the graph goes down.
f(x) is an even function so both ends of the graph go in the same direction.
is csc2 x + sec2x=1 an identity
Answer:
Step-by-step explanation:
=
sin
2
x
+
cos
2
x
sin
2
x
cos
2
x
=
1
Answer:
explanation is below happy to help ya!
Step-by-step explanation:
A real estate agent has 17 properties that she shows. She feels that there is a 40% chance of selling any one property during a week. The chance of selling any one property is independent of selling another property. Compute the probability of selling less than 5 properties in one week. Round your answer to four decimal places.
Answer:
0.126 = 12.6% probability of selling less than 5 properties in one week.
Step-by-step explanation:
For each property, there are only two possible outcomes. The chance of selling any one property is independent of selling another property. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
A real estate agent has 17 properties that she shows.
This means that [tex]n = 17[/tex]
She feels that there is a 40% chance of selling any one property during a week.
This means that [tex]p = 0.4[/tex]
Compute the probability of selling less than 5 properties in one week.
[tex]P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)[/tex]
In which
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{17,0}.(0.4)^{0}.(0.6)^{17} = 0.0002[/tex]
[tex]P(X = 1) = C_{17,1}.(0.4)^{1}.(0.6)^{16} = 0.0019[/tex]
[tex]P(X = 2) = C_{17,2}.(0.4)^{2}.(0.6)^{15} = 0.0102[/tex]
[tex]P(X = 3) = C_{17,3}.(0.4)^{3}.(0.6)^{14} = 0.0341[/tex]
[tex]P(X = 4) = C_{17,4}.(0.4)^{4}.(0.6)^{13} = 0.0796[/tex]
[tex]P(X < 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0002 + 0.0019 + 0.0102 + 0.0341 + 0.0796 = 0.126[/tex]
0.126 = 12.6% probability of selling less than 5 properties in one week.
What is the mean for the data set? 138,142,105,112,108,134,106,181,164,105
which graph represents this equation 5y=x+5
Answer: A
Step-by-step explanation: 5y=x+5 simplified to slope-intercept form is y = 1/5x + 1
The equation 5y = x + 5 is a linear equaton, and the graph A represents the equation 5y = x + 5
How to determine the graph?The equation is given as:
5y = x + 5
Make x = 0.
So, we have:
5y = 0 + 5
This gives
5y = 5
So, we have:
y = 1
The point is represented as (0,1)
Make y = 0.
So, we have:
5 * 0 = x + 5
This gives
0 = x + 5
So, we have:
x = -5
The point is represented as (-5,0)
This means that the graph of 5y = x + 5 passes through the points (0,1) and (-5,0)
From the list of options, only graph (A) passes through the points (0,1) and (-5,0)
Hence, the graph A represents the equation 5y = x + 5
Read more about linear equations at:
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Solve for e - 9e+4=−5e+14+13e
Answer:
e=1o ajjajajajajkananwnwnwhhshssh
Keep getting this wrong please help!!
Answer:
139.316148[tex]\pi[/tex]cubed inches
Step-by-step explanation:
If needed, use a calculator. Not sure if there's rounding for the final answer, so I will give you the precise answer.
1. Find the radius of the ball. If you know the circumference is 9.42 inches, divided it by 2, which the quotient is 4.71.
2. Remember the formula is V= [tex]\frac{4}{3} \pi r^3[/tex]
3. [tex]\frac{4}{3} \pi (4.71)^3[/tex]
4. [tex]\frac{4}{3} \pi (104.487111^3)[/tex]
5. [tex]4\pi (34.829037)[/tex]
6. 139.316148[tex]\pi[/tex]cubed inches
can someone help me?
For A: V = B · h V = 8 · 14 V = 784 units^3.
For B: V = B · h V = 9 · 9 V = 729 units^3.
For C: V = B · h V = 5 · 11 V = 220 units^3.
% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the number of U.S. adults who have very little confidence in newspapers is (a) exactly five, (b) at least six, and (c) less than four.
Complete Question:
41% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the number of U.S. adults who have very little confidence in newspapers is (a) exactly five, (b) at least six, and (c) less than four.
Answer:
a) P(exactly 5) = 0.209
b) P(at least six) = 0.183
c) P(less than four) = 0.358
Step-by-step explanation:
Sample size, n = 10
Proportion of adults that have very little confidence in newspapers, p = 41% p = 0.41
q = 1 - 0.41 = 0.59
This is a binomial distribution question:
[tex]P(X=r) = nCr p^{r} q^{n-r}[/tex]
a) P(exactly 5)
[tex]P(X=5) = 10C5 * 0.41^{5} 0.59^{10-5}\\P(X=5) = 10C5 * 0.41^{5} 0.59^{10-5}\\P(X=5) = 252 * 0.01159 * 0.072\\P(X=5) = 0.209[/tex]
b) P(at least six)
[tex]P(X \geq 6) = P(6) + P(7) + P(8) + P(9) + P(10)[/tex]
[tex]P(X\geq6) = (10C6 * 0.41^6*0.59^4) + (10C7*0.41^7*0.59^3) + (10C8*0.41^8*0.59^2) + (10C9 *0.41^9*0.59^1) + (10C10 *0.41^{10})\\P(X\geq6) = 0.1209 + 0.0480 + 0.0125 + 0.0019 + 0.0001\\P(X\geq6) = 0.183[/tex]
c) P(less than four)
[tex]P(X < 4) = 1 - [x \geq 4][/tex]
[tex]P(X<4)= 1 - [P(4) + P(5) + P(x \geq 6)][/tex]
[tex]P(X <4)= 1 - [(10C4*0.41^4*0.59^6) + 0.209 + 0.183]\\P(X <4)= 0.358[/tex]
What are the solution(s) to the quadratic equation 50 - x2 = 0?
x = +2,5
x = +63
x = +5,2
no real solution
Answer:
Step-by-step explanation:
Please use " ^ " for exponentiation: 50 - x^2 = 0.
Solving for x^2, we get: x^2 = 50
Taking the square root of both sides: x = ±√(50), or x = ±7.07
None of the three possible answers here is correct.
Petra jogs 5 miles in 40 minutes. At this rate, how long would it take her to jog 11 miles?
AHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH
Answer:
88 mins
Step-by-step explanation:
Divide 40 by 5 to get 8. That is your mile per minute ratio. Then multiply 11 by 8 to get the time of 11 miles!
In a data distribution, the first quartile, the median and the means are 30.8, 48.5 and 42.0 respectively. If the coefficient skewness is −0.38
a) What is the approximate value of the third quartile (Q3 ), correct to 2 decimal places.
b)What is the approximate value of the variance, correct to the nearest whole number
Answer:
a) The third quartile Q₃ = 56.45
b) The variance = 2633.31
Step-by-step explanation:
a) The coefficient of skewness formula is given as follows;
[tex]SK = \dfrac{Q_{3}+Q_{1}-2Q_{_{2}}}{Q_{3}-Q_{1}}[/tex]
Plugging in the values, we have;
[tex]-0.38 = \dfrac{Q_{3}+30.8-2 \times 48.5_{_{}}}{Q_{3}-30.8}[/tex]
Solving gives Q₃ = 56.45
b) To determine the variance, we use the skewness formula as follows;
[tex]SK_{p} = \dfrac{Mean-\left (3\times Median - 2\times Mean \right )}{\sigma } = \dfrac{3\times\left ( Mean - Median \right )}{\sigma }[/tex]
Plugging in the values, we get;
[tex]-0.38= \dfrac{42-\left (3\times 48.5- 2\times 42\right )}{\sigma } = \dfrac{-19.5}{\sigma}[/tex]
[tex]\therefore \sigma =\dfrac{-19.5}{-0.38} = 51.32[/tex]
The variance = σ² = 51.32² = 2633.31.
Rangers tagged and released 300 salmon into a Maine lake. A
month later, fishermen on the lake were surveyed. They reported
catching 80 salmon, 12 of which had tags. Using this sample,
estimate the salmon population in the lake.
Answer:
2,000 salmon
Step-by-step explanation:
The point estimate of the fraction of salmon that corresponds to the tagged Salmon released by the rangers, based on the fishermen report, is:
[tex]p = \frac{12}{80}\\p=0.15[/tex]
This means that the 300 salmon released into the lake correspond to 15% of the total salmon population. The estimate for the salmon population is:
[tex]P= \frac{300}{0.15}\\P=2,000\ salmon[/tex]
The estimate is 2,000 salmon.
Answer:
The estimate of the salmon population will be 2,000
Step-by-step explanation:
Rangers tagged and released 300 salmon into a Maine lake.
The fishermen reported catching 80 salmon of which 12 had tags
Thus we have 12/80 as a fraction of the tagged and released 300
12/80 = 0.15 which is about 15% of the population of tagged and released 300.
To then estimate the salmon population, we have 300/0.15 = 2,000
Find the GCF of -10c2d and 15 ca2
Answer:
5c
Step-by-step explanation:
5c is the only value you can take out of both factors
Is 343060 divisible by 8?
Answer:
no 8 is not divisible to 343060
The Wall Street Journal reports that 33% of taxpayers with adjusted gross incomes between $30,000 and $60,000 itemized deductions on their federal income tax return. The mean amount of deductions for this population of taxpayers was $16,642. Assume the standard deviation is σ = $2,400.
a. What is the probability that a sample of taxpayers from this income group who have itemized deductions will show a sample mean within $200 of the population mean for each of the following sample sizes: 20, 50, 100, and 500? (Round your answers to four decimal places.)
b. What is the advantage of a larger sample size when attempting to estimate the population mean?
a. A larger sample increases the probability that the sample mean will be a specified distance away from the population mean.
b. A larger sample increases the probability that the sample mean will be within a specified distance of the population mean.
c. A larger sample lowers the population standard deviation.
d. A larger sample has a standard error that is closer to the population standard deviation.
Answer:
(a) n : 20 50 100 500
P (-200 < X - μ < 200) : 0.2886 0.4444 0.5954 0.9376
(b) The correct option is (b).
Step-by-step explanation:
Let the random variable X represent the amount of deductions for taxpayers with adjusted gross incomes between $30,000 and $60,000 itemized deductions on their federal income tax return.
The mean amount of deductions is, μ = $16,642 and standard deviation is, σ = $2,400.
Assuming that the random variable X follows a normal distribution.
(a)
Compute the probability that a sample of taxpayers from this income group who have itemized deductions will show a sample mean within $200 of the population mean as follows:
For a sample size of n = 20[tex]P(\mu-200<X<\mu+200)=P(\frac{(16642-200)-16642}{2400/\sqrt{20}}<\frac{X-\mu}{\sigma/\sqrt{n}}<\frac{(16642+200)-16642}{2400/\sqrt{20}})[/tex]
[tex]=P(-0.37<Z<0.37)\\\\=P(Z<0.37)-P(Z<-0.37)\\\\=0.6443-0.3557\\\\=0.2886[/tex]
For a sample size of n = 50[tex]P(\mu-200<X<\mu+200)=P(\frac{(16642-200)-16642}{2400/\sqrt{50}}<\frac{X-\mu}{\sigma/\sqrt{n}}<\frac{(16642+200)-16642}{2400/\sqrt{50}})[/tex]
[tex]=P(-0.59<Z<0.59)\\\\=P(Z<0.59)-P(Z<-0.59)\\\\=0.7222-0.2778\\\\=0.4444[/tex]
For a sample size of n = 100[tex]P(\mu-200<X<\mu+200)=P(\frac{(16642-200)-16642}{2400/\sqrt{100}}<\frac{X-\mu}{\sigma/\sqrt{n}}<\frac{(16642+200)-16642}{2400/\sqrt{100}})[/tex]
[tex]=P(-0.83<Z<0.83)\\\\=P(Z<0.83)-P(Z<-0.83)\\\\=0.7977-0.2023\\\\=0.5954[/tex]
For a sample size of n = 500[tex]P(\mu-200<X<\mu+200)=P(\frac{(16642-200)-16642}{2400/\sqrt{500}}<\frac{X-\mu}{\sigma/\sqrt{n}}<\frac{(16642+200)-16642}{2400/\sqrt{500}})[/tex]
[tex]=P(-1.86<Z<1.86)\\\\=P(Z<1.86)-P(Z<-1.86)\\\\=0.9688-0.0312\\\\=0.9376[/tex]
n : 20 50 100 500
P (-200 < X - μ < 200) : 0.2886 0.4444 0.5954 0.9376
(b)
The law of large numbers, in probability concept, states that as we increase the sample size, the mean of the sample ([tex]\bar x[/tex]) approaches the whole population mean ([tex]\mu_{x}[/tex]).
Consider the probabilities computed in part (a).
As the sample size increases from 20 to 500 the probability that the sample mean is within $200 of the population mean gets closer to 1.
So, a larger sample increases the probability that the sample mean will be within a specified distance of the population mean.
Thus, the correct option is (b).
Using the normal distribution and the central limit theorem, it is found that:
a)
0.2886 = 28.86% probability that a sample size of 20 has a sample mean within $200 of the population mean.
0.4448 = 44.48% probability that a sample size of 50 has a sample mean within $200 of the population mean.
0.5934 = 59.34% probability that a sample size of 100 has a sample mean within $200 of the population mean.
0.9372 = 93.72% probability that a sample size of 500 has a sample mean within $200 of the population mean.
b)
b. A larger sample increases the probability that the sample mean will be within a specified distance of the population mean.
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
It measures how many standard deviations the measure is from the mean. After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.By the Central Limit Theorem, the standard deviation of the sampling distribution of sample sizes of n is: [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].Hence, the probability of sample of size n having a sample mean within 200 is the p-value of [tex]Z = \frac{200}{s}[/tex] subtracted by the p-value of [tex]Z = -\frac{200}{s}[/tex]
In this problem:
The standard deviation is [tex]\sigma = 2400[/tex].Item a:
For samples of 20, [tex]n = 20[/tex], and thus:
[tex]s = \frac{2400}{\sqrt{20}}[/tex]
Then
[tex]Z = \frac{200}{s} = \frac{200}{\frac{2400}{\sqrt{20}}} = 0.37[/tex]
Z = 0.37 has a p-value of 0.6443.Z = -0.37 has a p-value of 0.3557.0.6443 - 0.3557 = 0.2886.
0.2886 = 28.86% probability that a sample size of 20 has a sample mean within $200 of the population mean.
For samples of 50, [tex]n = 50[/tex], and thus:
[tex]s = \frac{2400}{\sqrt{50}}[/tex]
Then
[tex]Z = \frac{200}{s} = \frac{200}{\frac{2400}{\sqrt{50}}} = 0.59[/tex]
Z = 0.59 has a p-value of 0.7224.Z = -0.59 has a p-value of 0.2776.0.7224 - 0.2776 = 0.4448.
0.4448 = 44.48% probability that a sample size of 50 has a sample mean within $200 of the population mean.
For samples of 100, [tex]n = 100[/tex], and thus:
[tex]s = \frac{2400}{\sqrt{100}}[/tex]
Then
[tex]Z = \frac{200}{s} = \frac{200}{\frac{2400}{\sqrt{100}}} = 0.83[/tex]
Z = 0.83 has a p-value of 0.7967.Z = -0.83 has a p-value of 0.2033.0.7967 - 0.2033 = 0.5934
0.5934 = 59.34% probability that a sample size of 100 has a sample mean within $200 of the population mean.
For samples of 500, [tex]n = 500[/tex], and thus:
[tex]s = \frac{2400}{\sqrt{500}}[/tex]
Then
[tex]Z = \frac{200}{s} = \frac{200}{\frac{2400}{\sqrt{500}}} = 1.86[/tex]
Z = 1.86 has a p-value of 0.9686.Z = -1.86 has a p-value of 0.0314.0.9686 - 0.0314 = 0.9372
0.9372 = 93.72% probability that a sample size of 500 has a sample mean within $200 of the population mean.
Item b:
A larger sample size reduces the standard error, as [tex]s = \frac{\sigma}{\sqrt{n}}[/tex], that is, the standard error is inversely proportional to the square root of the sample size n, meaning that values are closer to the mean. Thus, the correct option is:
b. A larger sample increases the probability that the sample mean will be within a specified distance of the population mean.
A similar problem is given at https://brainly.com/question/24663213
The Office of Student Services at UNC would like to estimate the proportion of UNC's 28,500 students who are foreign students. In their random sample of 50 students, 4 are foreign students. Unknown to them, the proportion of all UNC students that are foreign students is 0.061. For each student, let x=1 if the student is foreign and let x=0 if the student is from the U.S.Find the mean and the standard deviation of the sampling distribution of the sample proportion for a sample of size 50.
Answer:
For the sampling distribution of the sample proportion for a sample of size 50, the mean is 0.061 and the standard deviation is 0.034.
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]
In this question:
[tex]p = 0.061, n = 50[/tex]
So
Mean:
[tex]\mu = p = 0.061[/tex]
Standard deviation:
[tex]s = \sqrt{\frac{0.061*0.939}{50}} = 0.034[/tex]
For the sampling distribution of the sample proportion for a sample of size 50, the mean is 0.061 and the standard deviation is 0.034.
coin is tossed and an eight-sided die numbered 1 through 8 is rolled. Find the probability of tossing a tail and then rolling a number greater than 5. The probability of tossing a tail and then rolling a number greater than 5 is nothing. (Round to three decimal places as needed.)
Answer:
The probability of tossing a tail and then rolling a number greater than 5 is 0.188
Step-by-step explanation:
Independent events:
If two events, A and B, are independent, we have that:
[tex]P(A \cap B) = P(A)*P(B)[/tex]
In this question:
The coin and the die are independent. So
Event A: Tossing a tail.
Event B: Rolling a number greater than 5.
Probability of tossing a tail:
Coin can be heads or tails(2 outcomes), so the probability of a tail is [tex]P(A) = \frac{1}{2} = 0.5[/tex]
Probability of rolling a number greater than 5:
8 numbers(1 through 8), 3 of which(6,7,8) are greater than 5. So the probability of rolling a number greater than 5 is [tex]P(B) = \frac{3}{8} = 0.375[/tex]
Probability of A and B:
[tex]P(A \cap B) = P(A)*P(B) = 0.5*0.375 = 0.188[/tex]
The probability of tossing a tail and then rolling a number greater than 5 is 0.188
The formula for the volume of a cube of side length L is: V = L X L X L What is the value of V when L= 5? *
Solution,
Length=5
Volume of cube=(L)^3
= 5*5*5
=125
hope it helps
Good luck on your assignment
You are given 2 to 1 odds against tossing three heads with three coins, meaning you win $2 if you succeed and you lose $1 if you fail. Find the expected value (to you) of the game. Would you expect to win or lose money in 1 game? In 100 games? Explain.
Answer:
- $0.625
Step-by-step explanation:
To win 3 heads must be obtained, the probability of this is:
p = (1/2) ^ 3 = 0.125
Now, let's review the other scenarios:
HHH -> 0.125
HHT
THH ----> p (2H, 1T) = 3 * 0125 = 0.375
HTH
HTT
TTH ----> p (1H, 2T) = 3 * 0125 = 0.375
THT
TTT -> 0.125
So the waiting value would be:
EV = 2 * 0.125 - 1 * 0.375 - 1 * 0.375 - 1 * 0.125
EV = - 0.625
That is to say that the waiting value is - $ 0.625
The outcome og 1 game cannot be predicted but 100 you loss because the expected value is negative
The waiting value would be "-$0.625", the expected value will be negative so the outcome of game 1 can't be predicted but the 100 games you lose.
According to the question,
Three heads must be obtained to win.The probability will be:
→ [tex]p= (\frac{1}{2} )^3[/tex]
[tex]= 0.125[/tex]
Now,
HHH:
→ 0.125
THH:
→ [tex]p(2H, 1T) = 3\times 0.125[/tex]
[tex]= 0.375[/tex]
TTH:
→ [tex]p(1H, 2T) = 3\times 0.125[/tex]
[tex]= 0.375[/tex]
hence,
The waiting value will be:
→ [tex]EV = 2\times 0.125 -1\times 0.375-1\times -0.375-1\times 0.125[/tex]
[tex]= -0.625[/tex]
Thus the above response is correct.
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What is 2 3/4 in a decimal format
Answer:
2.5
Step-by-step explanation:
Answer:
2.75
Step-by-step explanation:
2 x 4=8 + 3=11/4=2.75
Consider the following system of linear equations: Instructions: Solve the system by reducing its augmented matrix to reduced row echelon form (RREF). Yes, you must reduce it all the way to RREF. Write out the matrix at each step of the procedure, and be specific as to what row operations you use in each step. At the end of the procedure, clearly state the solution to the system outside of a matrix. 1. If the solution is unique, express the solution in real numbers. 3. If there are infinitely many solutions, express the solution in parameter(s). 3. If there is no solution, say so, and explain why.All of the following are possible ranks of a 4x3 matrix except:0123 4 How is the number of parameters in the general solution of a consistent linear system related to the rank of its coefficient matrix? Let r= number of rows in the coefficient matrix c= number of columns in the coefficient matrix p= number of parameters in the general solution R=rank of the coefficient matrix 1. R=p+r 2. R=C+p 3. R=r-p 4. R=C-p 5. R=p-
Answer:
Consider the following system of linear equations: 2 + 3y + 2z = 5 - 2x + y - z= -2 2x + 3z = 11 Instructions: • Solve the system by reducing its augmented matrix to reduced row echelon form (RREF). Yes, you must reduce it all the way to RREF. • Write out the matrix at each step of the procedure, and be specific as to what row operations you use in each step. • At the end of the procedure, clearly state the solution to the system outside of a matrix. • If the solution is unique, express the solution in real numbers. • If there are infinitely many solutions, express the solution in parameter(s). . If there is no solution, say so, and explain why.
All of the following are possible ranks of a 4x3 matrix EXCEPT O 1 2 3 4
How is the number of parameters in the general solution of a consistent linear system related to the rank of its coefficient matrix? Let r= number of rows in the coefficient matrix c= number of columns in the coefficient matrix p= number of parameters in the general solution R=rank of the coefficient matrix 1. R=p+r 2. R=C+p 3. R=r-p 4. R=C-p 5. R=p-r
Step-by-step explanation:
x + 3y +2z = 5
-2x + y - z = -2
2x + 3z = 11
Here,
[tex]A = \left[\begin{array}{ccc}1&3&2\\-2&1&-1\\2&0&3\end{array}\right][/tex]
[tex]B =\left[\begin{array}{ccc}5\\-2\\11\end{array}\right][/tex]
[tex]X=\left[\begin{array}{ccc}x\\y\\z\end{array}\right][/tex]
i.e AX=B
We can write as augmented matrix
[tex]\left[\begin{array}{ccc|c}1&3&2&5\\-2&1&-1&-2\\2&0&3&11\end{array}\right][/tex]
[tex]\frac{R_2\rightarrow R_2+2R_1}{R_3\rightarrow R_3-2R_1} \left[\begin{array}{ccc|c}1&3&2&5\\0&7&3&8\\0&-6&-1&1\end{array}\right][/tex]
[tex]\frac{R_3\rightarrow R_3+\frac{6R_2}{7} }{R_1\rightarrow R_1-\frac{3R_2}{7} } \left[\begin{array}{ccc|c}1&0&5/7&11/7\\0&7&3&8\\0&0&11/7&55/7\end{array}\right][/tex]
[tex]\frac{R_2\rightarrow\frac{R_2}{7}}{R_3\rightarrow\frac{7}{11}R_3} \left[\begin{array}{ccc|c}1&0&5/7&11/7\\0&1&3/7&8/7\\0&0&11/7&55/7\end{array}\right][/tex]
[tex]\frac{R_1\rightarrow R_1 -\frac{5}{7}R_3}{R_2\rightarrow R_2 -\frac{3}{7}R_3} \left[\begin{array}{ccc|c}1&0&0&-2\\0&1&0&-1\\0&0&1&5\end{array}\right][/tex]
Since Rank (A|B) = Rank (A) = 3 = number of variables
⇒ systems has unique solution and x = -2 , y = -1 , z = 5The reading speed of second grade students in a large city is approximately normal, with a mean of 90 words per
minute (wpm) and a standard deviation of 10 wpm. Complete parts (a) through (f).
-
(b) What is the probability that a random sample of 10 second grade students from the city results in a mean
reading rate of more than 96 words per minute?
The probability is ___
(Round to four decimal places as needed.)
Answer:
The probability that a random sample of 10 second grade students from the city results in a mean reading rate of more than 96 words per minute
P(x⁻>96) =0.0359
Step-by-step explanation:
Explanation:-
Given sample size 'n' =10
mean of the Population = 90 words per minute
standard deviation of the Population =10 wpm
we will use formula
[tex]Z = \frac{x^{-}-mean }{\frac{S.D}{\sqrt{n} } }[/tex]
Let X⁻ = 96
[tex]Z = \frac{96-90 }{\frac{10}{\sqrt{10} } }[/tex]
Z = 1.898
The probability that a random sample of 10 second grade students from the city results in a mean reading rate of more than 96 words per minute
[tex]P(X^{-}>x^{-} ) = P(Z > z^{-} )[/tex]
= 1- P( Z ≤z⁻)
= 1- P(Z<1.898)
= 1-(0.5 +A(1.898)
= 0.5 - A(1.898)
= 0.5 -0.4641 (From Normal table)
= 0.0359
Final answer:-
The probability that a random sample of 10 second grade students from
= 0.0359
Find the general solution of the simple homogeneous "system" below, which consists of a single linear equation. Give your answer as a linear combination of vectors. Let x2 and x3 be free variables. 3x1 - 6x2 9x3
Answer:
[tex]= \left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right] = x_2 \left[\begin{array}{ccc}2\\1\\0\end{array}\right] + x_3 \left[\begin{array}{ccc}-3\\0\\1\end{array}\right][/tex]
Step-by-step explanation:
Given: 3x1 - 6x2 + 9x3 = 0
x2 and x3 are free variables
We have:
3x1 = 6x2 - 9x3
divide all sides by 3, we have:
x1 = 2x2 - 3x3
Finding the general solution, we have:
[tex] \left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right] = \left[\begin{array}{ccc}2x_2 - 3x_3\\x_2\\x_3\end{array}\right] [/tex]
[tex] = \left[\begin{array}{ccc}2x_2\\x_2\\0\end{array}\right] + \left[\begin{array}{ccc}-3x_3\\0\\x_3\end{array}\right][/tex]
[tex]= x_2 \left[\begin{array}{ccc}2\\1\\0\end{array}\right] + x_3 \left[\begin{array}{ccc}-3\\0\\1\end{array}\right][/tex]
The general solution is
[tex]= \left[\begin{array}{ccc}x_1\\x_2\\x_3\end{array}\right] = x_2 \left[\begin{array}{ccc}2\\1\\0\end{array}\right] + x_3 \left[\begin{array}{ccc}-3\\0\\1\end{array}\right][/tex]
Which of the following are solutions to the quadratic equation x squared plus 8x=9
Answer:
X=1 x=-9
Step-by-step explanation:
(x^2)+8x=9
(x^2)+8x-9=0
(x-1)(x+9)=0
x=1 x=-9
The city manager made a scatter plot of the number of retail stores in a
city over many years. The trend line for the scatter plot has the equation y =
11x + 12, where y is the total number of stores and x is the number of years
after 1990. Predict how many retail stores there were in 1997.
Answer:
There were 89 retail stores in 1997.
Step-by-step explanation:
You know:
y = 11*x + 12
where y is the total number of stores and x is the number of years after 1990.
Being this a linear function (Linear functions are those functions that have the form y = mx + b as it happens in this case), to know the total number of stores and, then you must know the value of x, that is, the number of years after 1990, and replace that value in the given function.
To find out how many retail stores there were in 1997, then since 1990 7 years have passed, calculated by subtracting: 1997 - 1990 = 7
So, 7 is the number of years after 1990, then x=7
Replacing in the function y=11*x + 12 you get:
y=11*7+12
Solving:
y=77+12
y=89
There were 89 retail stores in 1997.
during each cycle, the velocity v (in meters per second) of a robotic welding device is given by v=9t-2/9+t^2, where t is time in seconds. find the expression for the displacement s (in meters) as a function of t if s=0 when t=0.
Answer:
[tex]d = \dfrac{9t^{2} }{2} - \dfrac{2}{9} t + \dfrac{t^3}{3}[/tex]
Step-by-step explanation:
Given the equation of velocity w.r.to time 't':
[tex]v=9t-\dfrac{2}{9}+t^2 ...... (1)[/tex]
Formula for Displacement:
[tex]Displacement = \text{velocity} \times \text{time}[/tex]
So, if we find integral of velocity w.r.to time, we will get displacement.
[tex]\Rightarrow \text{Displacement}=\int {v} \, dt[/tex]
[tex]\Rightarrow \int {v} \, dt = \int ({9t-\dfrac{2}{9}+t^2}) \, dt \\\Rightarrow \int{9t} \, dt - \int{\dfrac{2}{9}} \, dt + \int{t^2} \, dt\\\Rightarrow s=\dfrac{9t^{2} }{2} - \dfrac{2}{9} t + \dfrac{t^3}{3} + C ....... (1)[/tex]
Here, C is constant (because it is indefinite integral)
Formula for integration used:
[tex]1.\ \int({A+B}) \, dx = \int {A} \, dx + \int{B} \, dx \\2.\ \int({A-B}) \, dx = \int {A} \, dx - \int{B} \, dx \\3.\ \int{x^{n} } \, dx = \dfrac{x^{n+1}}{n+1}\\4.\ \int{C } \, dx = Cx\ \{\text{C is a constant}\}[/tex]
Now, it is given that s = 0, when t = 0.
Putting the values in equation (1):
[tex]0=\dfrac{9\times 0^{2} }{2} - \dfrac{2}{9}\times 0 + \dfrac{0^3}{3} + C\\\Rightarrow C = 0[/tex]
So, the equation for displacement becomes:
[tex]s=\dfrac{9t^{2} }{2} - \dfrac{2}{9} t + \dfrac{t^3}{3}[/tex]
3) For a certain good we have LaTeX: q=f\left(p\right)=200e^{-0.4p}q = f ( p ) = 200 e − 0.4 p.
a) Find the elasticity of demand at price p = $50.
b) At p = $50, is the demand elastic, inelastic, or does it have unit elasticity? Explain what this means for this product.
c) Find the elasticity of demand at price p = $20.
d) At p = $20, is the demand elastic, inelastic, or does it have unit elasticity? Explain what this means for this product.
Answer:
(a)20
(b)Elastic
(c)8
(d) Elastic
Step-by-step explanation:
Elasticity of demand(E) indicates the impact of a price change on a product's sales.
The general formula for an exponential demand curve is given as:
[tex]y=ae^{-bp}[/tex]
Given the demand curve formula
[tex]q=f\left(p\right)=200e^{-0.4p}[/tex]
The formula for Elasticity of demand, E
[tex]E = -\dfrac{p}{q}\dfrac{\text{d}q}{\text{d}p}[/tex]
(a)When Price, p = $50
p=50
[tex]q=200e^{-0.4*50}=200e^{-20}[/tex]
[tex]\dfrac{\text{d}q}{\text{d}p}=-0.4*200e^{-0.4p}=-80e^{-0.4p}[/tex]
Therefore:
[tex]E = -\dfrac{50}{200e^{-20}}*-80e^{-0.4*50}\\=\dfrac{1}{4e^{-20}}*80e^{-20}\\\\E=20[/tex]
(b)At p = $50, Since elasticity is greater than 1, the demand is elastic.
An elasticity value of 20 means that a 1% increase in price causes a 20% decrease in demand.
(c)At p=$20
p=20
[tex]q=200e^{-0.4*20}=200e^{-8}[/tex]
[tex]\dfrac{\text{d}q}{\text{d}p}=-0.4*200e^{-0.4p}=-80e^{-0.4p}[/tex]
Therefore:
[tex]E = -\dfrac{20}{200e^{-8}}*-80e^{-0.4*20}\\=\dfrac{1}{10e^{-20}}*80e^{-20}\\\\E=8[/tex]
(d)At p = $20, the demand is elastic.
An elasticity value of 8 means that a 1% increase in price causes a 8% decrease in demand.
