Answer:
[tex]10\frac{2}{5}[/tex]
Step-by-step explanation:
[tex]7\frac{3}{5} + x = 18\\[/tex]
transformation : [tex]7\frac{3}{5} = \frac{(5*7) + 3}{5} = \frac{38}{5}[/tex]
[tex]\frac{38}{5} + x = 18\\\\x = 18 - \frac{38}{5} = \frac{(5*18) - 38}{5} = \frac{52}{5}[/tex]
[tex]\frac{52}{5} = 10\frac{2}{5}[/tex]
Answer:
[tex]10\frac{2}{5}[/tex]
Step-by-step explanation:
What should be add to [tex]7\frac{3}{5}[/tex] to get [tex]18[/tex] ?
[tex]18-7\frac{3}{5}=\\ \\=17\frac{5}{5}-7\frac{3}{5}\\\\\\=(17-7)+(\frac{5}{5}- \frac{3}{5})\\ \\=10+\frac{2}{5}\\ \\=10\frac{2}{5}[/tex]
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 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.
Miles draws three cards at random from a standard deck of 52 cards, without replacement.
(a) Find the probability that all three cards are red.
(b) Find the probability that all three cards have the same rank.
Answer:
P( red, red, red no replacement ) = 2/17
P( 3 cards with same rank, no replacement) = 1/425
Step-by-step explanation:
There are 52 cards, 26 are red
P( red) = red/total =26/ 52 = 1/2
Now draw the second card without replacement
There are 51 cards, 25 are red
P( red) = red/total =25/51
Now draw the third card without replacement
There are 50 cards, 24 are red
P( red) = red/total =24/50 = 12/25
P( red, red, red no replacement ) = 1/2 * 25/51 * 12/25 = 2/17
P ( all have the same rank)
There are 52 cards, 4 of each rank
We don't care what the first rank is, now the second and third draw have to have the same rank as the first
P( rank) = 1/1
Now draw the second card without replacement
There are 51 cards, 3 are left with the same rank
P( same rank) = cards with same rank/total = 3/51
Now draw the third card without replacement
There are 50 cards, 2 are left with the same rank
P( same rank) = cards with same rank/total = 2/50 = 1/25
P( 3 cards with same rank, no replacement) = 1 * 3/51*1/25 = 1/425
BACTROBAN ointment contains 2% w/w mupirocin. How many grams of a polyethylene glycol ointment base must be mixed with the contents of a 22-g tube of the BACTROBAN ointment to prepare one having a concentration of 5 mg/g?
Answer:
66g of polyethylene ointment base!
Suppose that two people standing 2 miles apart both see the burst from a fireworks display. After a period of time, the first person standing at point A hears the burst. Six seconds later, the second person standing at point B hears the burst. If the person at point B is due west of the person at point A and if the display is known to occur due north of the person at point A, where did the fireworks display occur? Note that sound travels at 1100 feet per second.
Answer: approximately 5148 feet directly north of point A
This is equivalent to 0.975 miles
==============================================================
Explanation:
Check out figure 1 (attached image below) to see how the drawing is set up.
Points A and B are separated by 2 miles, or 10560 feet. Use the conversion factor 1 mile = 5280 feet.
Because sound travels at roughly 1100 feet per second (assuming all of the conditions are right), this means that after x seconds, the sound wave has traveled a total distance of 1100x feet. This is the distance from A to C. Point C is the firework location.
The distance from B to C is 1100(x+6) feet because it takes 6 more seconds for the soundwave to travel from C to B, compared from C to A.
Note that triangle ABC is a right triangle. The 90 degree angle is at point A.
We can use the pythagorean theorem to solve for x
See figure 2 in the attached images to see the steps on solving for x. The numbers get quite big, though in the end x is fairly small. We end up with x = 4.68, which is approximate due to the 1100 figure being approximate.
This means the sound wave takes about 4.68 seconds to go from point C to point A.
With this x value, we can compute the expression 1100x to get
1100*x = 1100*4.68 = 5,148 feet
To convert this to miles, divide by 5280
(5148)/(5280) = 0.975
Therefore,
5148 feet = 0.975 miles
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.
The total cost (in dollars) for a company to manufacture and sell x items per week is C = 40 x + 640 , whereas the revenue brought in by selling all x items is R = 66 x − 0.2 x 2 . How many items must be sold to obtain a weekly profit of $ 200 ?
Answer:
The company must sell 60 or 70 items to obtain a weekly profit of 200.
Step-by-step explanation:
The profit is the difference between the revenue and the cost of a given task, therefore:
[tex]\text{profit} = R - C\\\text{profit} = 66*x - 0.2*x^2 - (40*x + 640)\\\text{profit} = 66*x - 0.2*x^2 - 40*x - 640\\\text{profit} = - 0.2*x^2 + 26*x - 640[/tex]
To have a profit of 200, we need to sell:
[tex]-0.2*x^2 + 26*x - 640 = 200\\-0.2*x^2 + 26*x -840 = 0\text{ } *\frac{-1}{0.2}\\x^2 -130 + -4200 = 0\\x_{1,2} = \frac{-(-130) \pm \sqrt{(-130)^2 - 4*1*(-4200)}}{2*1}\\x_{1,2} = \frac{130 \pm \sqrt{16900 + 16800}}{2}\\x_{1,2} = \frac{130 \pm \sqrt{100}}{2}\\x_{1,2} = \frac{130 \pm 10}{2}\\x_{1} = \frac{130 + 10}{2} = \frac{140}{2} = 70\\ x_{2} = \frac{130 - 10}{2} = \frac{120}{2} = 60[/tex]
The company must sell 60 or 70 items to obtain a weekly profit of 200.
The congressional committees on mathematics and computer science are made up of five representatives each, and a congressional rule is that the two committees must be disjoint. If there are 385 members of congress, how many ways could the committees be selected?
Answer:
The committes can be selected in [tex]1.752507297 \times 10^{19}[/tex] ways
Step-by-step explanation:
The order in which the members are chosen to the committee is not important. So we use the combinations formula to solve this question.
Combinations formula:
[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]
The two committees must be disjoint.
This means that a person cannot be part of both committes.
If there are 385 members of congress, how many ways could the committees be selected?
Since the committes are disjoint, 5(math) + 5(computer science) = 10 people will be chosen from the set of 385. So
[tex]C_{385,10} = \frac{385!}{10!(385-10)!} = 1.752507297 \times 10^{19}[/tex]
The committes can be selected in [tex]1.752507297 \times 10^{19}[/tex] ways
Is 343060 divisible by 8?
Answer:
no 8 is not divisible to 343060
The 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
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]
What is six thousandths as a number?
Answer:
6000
Step-by-step explanation:
6 six
60 sixty
600 six hundred
6000 six thousand
//have a great day//
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.
Learn more:
https://brainly.com/question/13753898
I have a question about this.
What is the mean for the data set? 138,142,105,112,108,134,106,181,164,105
Which numbers are the extremes of the proportion shown below
Answer:
3 and 8 are the extremes of the given proportion (first listed answer option)
Step-by-step explanation:
In a proportion of the type:
[tex]\frac{A}{B} =\frac{C}{D}[/tex]
the values "A" and "D" are called "extremes of the proportion, while "B" and "C" are called the "means" of the proportion.
This comes from writing it as: A : B = C : D
where we observe that A and D are the more "extreme" values being further away from the equal sign than the quantities B and C which are more towards the middle (center of the proportion).
So in your case, the numbers 3 and 8 are the extremes
Answer:
3 and 8
Step-by-step explanation:
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:
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.
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.
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
What is the value of x?
Answer:
Oioj[
Step-by-step explanation:
Step-by-step explanation:
Write the equation to solve it in order to know the exact value of X.
In how many ways you can put 20 books on 50 shelves so that there is not more than one book on a shelf?
9514 1404 393
Answer:
47129212243960 ≈ 4.713×10^13 if order doesn't matter114660755112113373922453094400000 ≈ 1.147×10^32 if it doesStep-by-step explanation:
If order doesn't matter, the number is the combinations of 50 things taken 20 at a time:
50C20 = 47129212243960 ≈ 4.713×10^13
If order does matter, then the number of ways is this value multiplied by 20!:
50P20 = 114660755112113373922453094400000 ≈ 1.147×10^32
_____
nCk = n!/(k!(n-k)!)
nPk = n!/(n-k)!
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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3 lines intersect. A horizontal line with points G, N, K intersects a vertical line with points J, N, M at point N and forms a right angle. The third line contains points H, N, L and intersects the other lines at N. It is diagonal and cuts through angles J N G and M N K. Which statements are true regarding the diagram? AngleGNH and AngleHNJ are complementary. AngleJNK and AngleKNL are supplementary. AngleKNL and AngleLNM are complementary. mAngleHNK + mAngleKNL = 180° mAngleMNG + mAngleGNH = 90°
Answer:
A, C, D
Step-by-step explanation:
Answer: A , C , D
Step-by-step explanation: edge2022
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
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
An experiment was done to look at whether there is an effect of the number of hours spent practising a musical instrument and gender on the level of musical ability. A sample of 30 (15 men and 15 women) participants who had never learnt to play a musical instrument before were recruited. Participants were randomly allocated to one of three groups that varied in the number of hours they would spend practising every day for 1 year (0 hours, 1 hours, 2 hours). Men and women were divided equally across groups. All participants had a one-hour lesson each week over the course of the year, after which their level of musical skill was measured on a 10-point scale ranging from 0 (you can’t play for toffee) to 10 (‘Are you Mozart reincarnated?’). An ANOVA was conducted on these data, and the results revealed significant main effects of gender and the number of hours spent practising. Which of the following contrasts could we use to break down the effect of gender?
A- Helmert contratst
B- Simple contrast
C- Repeated contrast
D- None
Answer: Repeated contrast
Step-by-step explanation:
In the two way ANOVA that was performed above, there were 3 groups of which 30 people (15 men and 15 women) participants who had never learnt to play a musical instrument before were recruited and divided equally.
The ANOVA is of repeated measures. There is within group effect, between group effect and interaction effect. The result also shows significant main effect for the gender and the hours used practicing. Hence, to breakdown the effect of gender, the repeated contrast method will be utilized. The repeated contrast method compares the mean of every level with the succeeding level except the last level.
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.
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