y
6
Which points are located on this graph? Check all that
apply.
4
(1,1)
3
B (2,4)
2
(-3,0)
(0, 3)
(3,5)
-5 -4 -3 -2 -1
2 3
4
5
-2
(1,4)
-3
-4
-5
plssssss hellpppp
Answer:
(-3,0) (0,3) (1,4)
Step-by-step explanation:
Answer:
the two you selected and the last choice
(-3, 0)
(0, 3)
(1, 4)
Step-by-step explanation:
A snack stand sells bottles of water for $1.35 each . Sam needs to buy as many bottles as he can. He can spend no more than $20. How much money will Sam have left after he buys the bottles of water?
Answer: 1.1
Step-by-step explanation:
If Sam buys 14 water bottles, he will have spent 18.9, so when you subtract 20 (the limit) from 18.9 (how much money he'll spend on the water bottles) you get 1.1
An investment of $100 comma 000 was made by a business club. The investment was split into three parts and lasted for one year. The first part of the investment earned 8% interest, the second 6%, and the third 9%. Total interest from the investments was $ 7380. The interest from the first investment was 2 times the interest from the second. Find the amounts of the three parts of the investment.
Answer:
$54000 was invested at 8% interest$36000 was invested at 6% interest$10000 was invested at 9% interestStep-by-step explanation:
Total Investment = $100,000
The investment was split into three parts (say x, y and z)
x+y+z=100000Simple Interest = Principal X Rate/100 X Time
The first part of the investment earned 8% interest
Interest on the first part = 0.08x
The second part of the investment earned 6% interest
Interest on the second part = 0.06y
The third part of the investment earned 9% interest
Interest on the third part = 0.09z
Total interest from the investments = $ 7380.
Therefore:
0.08x+0.06y+0.09z=7380
The interest from the first investment was 2 times the interest from the second.
0.08x=2 X 0.06y
0.08x=0.12y
Substituting we have:
0.12y+0.06y+0.09z=7380
0.18y+0.09z=7380
From 0.08x=0.12y
x=1.5y
Substituting x=1.5y into x+y+z=100000
1.5y+y+z=100000
2.5y+z=100000
z=100000-2.5y
Substituting z=100000-2.5y into 0.18y+0.09z=7380
0.18y+0.09(100000-2.5y)=7380
0.18y+9000-0.225y=7380
-0.045y=-1620
Divide both sides by -0.045
y=36000
Recall: x=1.5y
x=1.5 X 36000
x =54000
x+y+z=100000
54000+36000+z=100000
z=100000-(54000+36000)
z=10,000
Therefore:
$54000 was invested at 8% interest$36000 was invested at 6% interest$10000 was invested at 9% interestSelect one: A. ∠T ≅ ∠F B. ∠T ≅ ∠D C. ∠J ≅ ∠F D. ∠J ≅ ∠D
Answer:
B
Step-by-step explanation:
Since you are given line RT and ND, then all you need to do by ASA is the make sure that the angles are the two endpoints are congruent. Since the problem already gave you R and N, then all thats left is to relate the other two endpoints, namely, T and D
What is the value of b^2- 4ac for the following equation? x(x+8)=9
Answer:
hope this is correct
5. A meteorologist measured the average rainfall received in cities A and
B. Both cities received 11 inches of rainfall in total. While City A received x
inches of rain, City B experienced three times the amount of rainfall than
City A. Find the number of inches of rain City A received.
Answer:
2.75 inches
Step-by-step explanation:
City A received x inches of rain.City B experienced three times the amount of rainfall than City A, therefore :
City B received 3x inches of rain.Since both cities received 11 inches of rainfall in total.
We have that:
x+3x=11
4x=11
Divide both sides by 4
x=2.75 Inches
Therefore, we City A received 2.75 inches of rain.
Notice that the alternative hypothesis is a two-tailed test. Suppose ttest_ind method from scipy module is used to perform the test and the output is (-1.99, 0.0512). What is the P-value for this hypothesis test
Answer:
The p-value of the test is 0.0512.
Step-by-step explanation:
The p-value of a test is well-defined as per the probability, [under the null hypothesis (H₀)], of attaining a result equivalent to or more extreme than what was the truly observed value of the test statistic.
In this case the output of the t-test_ind method from scipy module is provided as follows:
Output = (-1.99, 0.0512)
The first value within the parentheses is the test statistic value.
So the test statistic value is, -1.99.
And the second value within the parentheses is the p-value of the test.
So the p-value of the test is 0.0512.
i need halp pls im confused and it would help me out
lateral area
surface area
volume
Step-by-step explanation:
lateral area=area of the full triangle
surface area=area of the four triangles + area of the square
This table gives a few (x,y) pairs of a line in the coordinate plane.
Answer:
-26
Step-by-step explanation:
y=mx+c
m= (y2 - y1 ) / (x2-x1)
m= (10-1) / (48 -36)
m = 0.75
10 = 0.75(48) + c
c = 10 - 36
c = -26
Assume the readings on thermometers are normally distributed with a mean of 0degreesC and a standard deviation of 1.00degreesC. Find the probability that a randomly selected thermometer reads between negative 2.05 and negative 1.49 and draw a sketch of the region.
Answer:
[tex]P(-2.05<X<-1.49)=P(\frac{-2.05-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{-1.49-\mu}{\sigma})=P(\frac{-2.05-0}{1}<Z<\frac{-1.49-0}{1})=P(-2.05<z<-1.49)[/tex]
And we can find this probability with this difference
[tex]P(-2.05<z<-1.49)=P(z<-1.49)-P(z<-2.05)=0.068- 0.0202= 0.0478[/tex]
And we can see the figure in the plot attached.
Step-by-step explanation:
Let X the random variable that represent the redings on thermometers of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(0,1)[/tex]
Where [tex]\mu=0[/tex] and [tex]\sigma=1[/tex]
We are interested on this probability
[tex]P(-2.05<X<-1.49)[/tex]
We can use the z score formula given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
Using this formula we got:
[tex]P(-2.05<X<-1.49)=P(\frac{-2.05-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{-1.49-\mu}{\sigma})=P(\frac{-2.05-0}{1}<Z<\frac{-1.49-0}{1})=P(-2.05<z<-1.49)[/tex]
And we can find this probability with this difference
[tex]P(-2.05<z<-1.49)=P(z<-1.49)-P(z<-2.05)=0.068- 0.0202= 0.0478[/tex]
And we can see the figure in the plot attached.
Juanita goes shopping in Times Square, where T-shirts are 20% off today. If she buys one in the next 10 minutes, she can get an extra 10% off the sale price. How much would Juanita pay for a T-shirt right now? *
Answer:
72%
Step-by-step explanation:
As the T-shirts have 20% off and Juanita would get an extra 10% off the sale price, the additional discount would be over the price with the 20% off included. This means that the sale price is the 80% of the normal price and an additional 10% over that 80% would be:
80*10%= 8
80%-8%= 72%
This means that Juanita would have to pay the 72% percent of the regular price of the T-shirt as the total discount would be 28% off.
1. (a) The life time of a certain brand of bulbs produced by a company is normally distributed, with mean 210 hours and standard deviation 56 hours. What is the probability that a bulb picked at random from this company’s products will have a life time of:
(i) (ii) (iii) at least 300 hours,
at most 100 hours, between 150 and 250 hours.
(b) In a contest, two friends, Kofi and Mensah were asked to solve a problem. The probability that Kofi will solve it correctly is and the probability that Mensah
will solve it correctly is . Find the probability that neither of them solved it correctly.
2. Suppose that the random variable, X, is a number on the biased die and the p.d.f. of X is as shown below;
X
1
2
3
4
5
6
P(X=x)
1/6
1/6
1/5
k
1/5
1/6
a) Find;
(i) (ii) (iii) (iv) (v) the value of k. E(X)
E(X2) Var(X) P(1 £X <5)
b) If events A and B are such that they are independent, and P(A) = 0.3 with P(B) = 0.5;
i. ii. Find P(A n B) and P(AUB)
Are A and B mutually exclusive? Explain.
c) In how many ways can the letters of the word STATISTICS be arranged?
Answer:
See explanation
Step-by-step explanation:
Q1)a
- Denote a random variable ( X ) as the life time of a brand of bulb produced.
- The given mean ( μ ) = 210 hrs and standard deviation ( σ ) = 56 hrs. The distribution is symbolized as follows:
X ~ Norm ( 210 , 56^2 )
i) The bulb picked to have a life time of at least 300 hours.
- We will first standardize the limiting value of the RV ( X ) and determine the corresponding Z-score value:
P ( X ≥ x ) = P ( Z ≥ ( x - μ ) / σ )
P ( X ≥ 300 ) = P ( Z ≥ ( 300 - 210 ) / 56 )
P ( X ≥ 300 ) = P ( Z ≥ 1.607 )
- Use the standard normal look-up table for limiting value of Z-score:
P ( X ≥ 300 ) = P ( Z ≥ 1.607 ) = 0.054 .. Answer
ii) The bulb picked to have a life time of at most 100 hours.
- We will first standardize the limiting value of the RV ( X ) and determine the corresponding Z-score value:
P ( X ≤ x ) = P ( Z ≤ ( x - μ ) / σ )
P ( X ≤ 100 ) = P ( Z ≤ ( 100 - 210 ) / 56 )
P ( X ≤ 100 ) = P ( Z ≤ -1.9643 )
- Use the standard normal look-up table for limiting value of Z-score:
P ( X ≤ 100 ) = P ( Z ≤ -1.9643 ) = 0.0247 .. Answer
iii) The bulb picked to have a life time of between 150 and 250 hours.
- We will first standardize the limiting value of the RV ( X ) and determine the corresponding Z-score value:
P ( x1 ≤ X ≤ x2 ) = P ( ( x1 - μ ) / σ ≤ Z ≤ ( x2 - μ ) / σ )
P ( 150 ≤ X ≤ 250 ) = P ( ( 150 - 210 ) / 56 ≤ Z ≤ ( 250 - 210 ) / 56 )
P ( 150 ≤ X ≤ 250 ) = P ( -1.0714 ≤ Z ≤ 0.71428 )
- Use the standard normal look-up table for limiting value of Z-score:
P ( 150 ≤ X ≤ 250 ) = P ( -1.0714 ≤ Z ≤ 0.71428 ) = 0.6205 .. Answer
Q1)b
- Denote event (A) : Kofi solves the problem correctly. Then the probability of him answering successfully is:
p ( A ) = 0.25
- Denote event (B) : Menesh solves the problem correctly. Then the probability of him answering successfully is:
p ( B ) = 0.4
- The probability that neither of them answer the question correctly is defined by a combination of both events ( A & B ). The two events are independent.
- So for independent events the required probability can be stated as:
p ( A' & B' ) = p ( A' ) * p ( B' )
p ( A' & B' ) = [ 1 - p ( A ) ] * [ 1 - p ( B ) ]
p ( A' & B' ) = [ 1 - 0.25 ] * [ 1 - 0.4 ]
p ( A' & B' ) = 0.45 ... Answer
Q2)a
- A discrete random variable X: defines the probability of getting each number on a biased die.
- From the law of total occurrences. The sum of probability of all possible outcomes is always equal to 1.
∑ p ( X = xi ) = 1
p ( X = 1 ) + p ( X = 2 ) + p ( X = 3 ) + p ( X = 4 ) + p ( X = 5 ) + p ( X = 6 )
1/6 + 1/6 + 1/5 + k + 1/5 + 1/6 = 1
k = 0.1 ... Answer
- The expected value E ( X ) or mean value for the discrete distribution is determined from the following formula:
E ( X ) = ∑ p ( X = xi ) . xi
E ( X ) = (1/6)*1 + (1/6)*2 + (1/5)*3 + (0.1)*4 + (1/5)*5 + (1/6)*6
E ( X ) = 3.5 .. Answer
- The expected-square value E ( X^2 ) or squared-mean value for the discrete distribution is determined from the following formula:
E ( X^2 ) = ∑ p ( X = xi ) . xi^2
E ( X^2 ) = (1/6)*1 + (1/6)*4 + (1/5)*9 + (0.1)*16 + (1/5)*25 + (1/6)*36
E ( X^2 ) = 15.233 .. Answer
- The variance of the discrete random distribution for the variable X can be determined from:
Var ( X ) = E ( X^2 ) - [ E ( X ) ] ^2
Var ( X ) = 15.2333 - [ 3.5 ] ^2
Var ( X ) = 2.9833 ... Answer
- The cumulative probability of getting any number between 1 and 5 can be determined from the sum:
P ( 1 < X < 5 ) = P ( X = 2 ) + P ( X = 3 ) + P ( X = 4 )
P ( 1 < X < 5 ) = 1/6 + 1/5 + 0.1
P ( 1 < X < 5 ) = 0.467 ... Answer
Q2)b
- Two independent events are defined by their probabilities as follows:
p ( A ) = 0.3 and p ( B ) = 0.5
- The occurrences of either event does not change alter or affect the occurrences of the other event; hence, independent.
- For the two events to occur simultaneously at the same time:
p ( A & B ) = p ( A )* p ( B )
p ( A & B ) = 0.3*0.5
p ( A & B ) = 0.15 ... Answer
- For either of the events to occur but not both. From the comparatively law of two independent events A and B we have:
p ( A U B ) = p ( A ) + p ( B ) - 2*p ( A & B )
p ( A U B ) = 0.3 + 0.5 - 2*0.15
p ( A U B ) = 0.5 ... Answer
- Two mutually exclusive events can-not occur simultaneously; hence, the two events are not mutually exclusive because:
p ( A & B ) = 0.15 ≠ 0
Q2)c
- The letters of the word given are to be arranged in number of different ways as follows:
STATISTICS
- Number of each letters:
S : 3
T : 3
A: 1
I: 2
C: 1
- 10 letters can be arranged in 10! ways.
- However, the letters ( S and T and I ) are repeated. So the number of permutations must be discounted by the number of each letter is repeated as follows:
[tex]\frac{10!}{3!3!2!} = \frac{3628800}{72} = 50,400[/tex]
- So the total number of ways the word " STATISTICS " can be re-arranged is 50,400 without repetitions.
In a game of poker a player receives a subset of 5 cards from a standard deck of 52 cards. There are four suites (clubs, spades, hearts, diamonds) with 13 cards in each suite. What is the probability your hand contains exactly 3 hearts (out of 5 cards)?
Answer:
8.15% probability your hand contains exactly 3 hearts
Step-by-step explanation:
A probability is the number of desired outcomes divided by the number of total outcomes.
The order in which the cards are chosen 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]
What is the probability your hand contains exactly 3 hearts (out of 5 cards)?
We have 52 cards in total.
13 are clubs, 13 are spades, 13 are hearts, 13 are diamonds.
Desired outcomes:
3 hearts, from a set of 13.
Other 2 cards, from a set of 52-13 = 39.
So
[tex]D = C_{13,3}*C_{39,2} = \frac{13!}{3!(13-3)!}*\frac{39!}{2!(39-2)!} = 211926[/tex]
Total outcomes:
5 cards, from a set of 52. So
[tex]T = C_{52,5} = \frac{52!}{5!47!} = 2598960[/tex]
Probabilities:
[tex]p = \frac{D}{T} = \frac{211926}{2598960} = 0.0815[/tex]
8.15% probability your hand contains exactly 3 hearts
Todd is 3 years older than his brother Jack. If Jack is x years old and Todd is y years
old, write a rule that relates their ages over time. When Jack is 28 years old, how old will Todd be?
Answer:
1) y = x + 3
2) 31 years old
Step-by-step explanation:
Jack is x years old
Todd is y years old
Todd is 3 years older than his brother Jack => y = x + 3
When Jack is 28 years old, Todd is y = 28 + 3 = 31 years old
Apple weights in an orchard are normally distributed. From a sample farmer Fred determines the mean weight of a box of apples to be 270 oz. with a standard deviation of 10 oz. He wonders what percent of the apple boxes he has grouped for sale will have a weight less than 255 oz.
Answer:
2.07 is the answer as of e.2020
Step-by-step explanation:
Sam is rowing a boat away from a dock. The graph shows the relationship
between time and Sam's distance from the dock. Evaluate the function for an
input of 6.
Distance from Dock
Distance (meters)
110
100
90
80
70
50
50
40
10
20
0 1 2 3 4 5 6 7 8 9 10 11
Times (minutes)
O A. After 60 minutes, Sam is 6 meters from the dock
O B. After 6 minutes, Sam is 60 meters from the dock.
O c. After 60 minutes, Sam is 60 meters from the dock.
Answer: after 6 mins, sam is 60 meters from the dock
Step-by-step explanation:
Answer:
after 6 mins, sam is 60 meters from the dock
Step-by-step explanation:
eu (european union) countries report that 46% of their labor force is female. The United Nations wnats to determine if the percentage of femailes in the U.S. labor force is the same. Based on sample, representatives from the United States department of labor find that the 95% confidence interval for the proportion of females in the U.S. labor force is .357 to .443. if the department of labores wishes to tighten it's interval they should
Answer:
Step-by-step explanation:
The question is incomplete. The complete question is:
EU (European Union) countries report that 46% of their labor force is female. The United Nations wants to determine if the percentage of females in the U.S. labor force is the same. Based on a sample of 500 employment records, representatives from the United States Department of Labor found that the 95% confidence interval for the proportion of females in the U.S. labor force is 0.357 to 0.443. If the Department of Labor wishes to tighten its interval, they should:
A. increase the confidence level
B. decrease the sample size
C. increase the sample size
D. Both A and B
E. Both A and C
Solution:
Confidence interval for population proportion is written as
Sample proportion ± margin of error
Where sample proportion is the point estimate for the population proportion.
Margin of error = z × √pq/n
The z score for 95% confidence level is 1.96
p = 46/100 = 0.46
q = 1 - p = 1 - 0.46
q = 0.54
n = 500
Margin of error = 1.96√0.46 × 0.54/500 = 0.044
To tighten it's interval, the margin of error needs to be reduced.
If we increase the confidence level, say to 99%, z = 2.58
Then
Margin of error = 2.58√0.46 × 0.54/500 = 0.058
It increased
Also, If we increase the sample size, say to 700, then
Margin of error = 1.96√0.46 × 0.54/700 = 0.037
It has reduced
Therefore, the correct options is
C. increase the sample size
A local movie theater increased the price of
admission by 20%. Tickets had sold for $5.25.
What is the current ticket price?
Answer:
0.26
Step-by-step explanation:
10. A rectangular tank has length 4 m, width
12 m and height 3.5 m.
i) If the tank is filled with water to 2/3
of its capacity, calculate the volume
of the water in the tank.
Answer:
112m³
Step-by-step explanation:
VOLUME=L × W × H
= 4m × 12m × 3.5m
= 168m³
THE TANK IS FILLED TO 2/3 OF IT'S CAPACITY
= 2/3 × 168
= 112m³.
Jennifer started a lawn mowing business and borrowed $85 to buy materials. Nicholas started a babysitting business and borrowed $60 to buy materials. Which statement correctly shows who has the greater debt?
Answer:
Jennifer because she would make less but spent more on supplies
Step-by-step explanation:
Answer: The answer is D! Jennifer, |-85| > |-60|
Step-by-step explanation: I did the quiz hope everyone does great ^^
An industrial sewing machine uses ball bearings that are targeted to have a diameter of 0.75 inch. The lower and upper specification limits under which the ball bearings can operate are 0.74 inch and 0.76 inch, respectively. Past experience has indicated that the actual diameter of the ball bearings is approximately normally distributed, with a mean of 0.753 inch and a standard deviation of 0.004 inch. What is the probability that a ball bearing is:___________.
a. between the target and the actual mean?
b. between the lower specification limit and the target?
c. above the upper specification limit?d. below the lower specification limit?
Answer:
(a) Probability that a ball bearing is between the target and the actual mean is 0.2734.
(b) Probability that a ball bearing is between the lower specification limit and the target is 0.226.
(c) Probability that a ball bearing is above the upper specification limit is 0.0401.
(d) Probability that a ball bearing is below the lower specification limit is 0.0006.
Step-by-step explanation:
We are given that an industrial sewing machine uses ball bearings that are targeted to have a diameter of 0.75 inch. The lower and upper specification limits under which the ball bearings can operate are 0.74 inch and 0.76 inch, respectively.
Past experience has indicated that the actual diameter of the ball bearings is approximately normally distributed, with a mean of 0.753 inch and a standard deviation of 0.004 inch.
Let X = diameter of the ball bearings
SO, X ~ Normal([tex]\mu=0.753,\sigma^{2} =0.004^{2}[/tex])
The z-score probability distribution for normal distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma} } }[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean = 0.753 inch
[tex]\sigma[/tex] = standard deviation = 0.004 inch
(a) Probability that a ball bearing is between the target and the actual mean is given by = P(0.75 < X < 0.753) = P(X < 0.753 inch) - P(X [tex]\leq[/tex] 0.75 inch)
P(X < 0.753) = P( [tex]\frac{X-\mu}{\sigma} } }[/tex] < [tex]\frac{0.753-0.753}{0.004} } }[/tex] ) = P(Z < 0) = 0.50
P(X [tex]\leq[/tex] 0.75) = P( [tex]\frac{X-\mu}{\sigma} } }[/tex] [tex]\leq[/tex] [tex]\frac{0.75-0.753}{0.004} } }[/tex] ) = P(Z [tex]\leq[/tex] -0.75) = 1 - P(Z < 0.75)
= 1 - 0.7734 = 0.2266
The above probability is calculated by looking at the value of x = 0 and x = 0.75 in the z table which has an area of 0.50 and 0.7734 respectively.
Therefore, P(0.75 inch < X < 0.753 inch) = 0.50 - 0.2266 = 0.2734.
(b) Probability that a ball bearing is between the lower specification limit and the target is given by = P(0.74 < X < 0.75) = P(X < 0.75 inch) - P(X [tex]\leq[/tex] 0.74 inch)
P(X < 0.75) = P( [tex]\frac{X-\mu}{\sigma} } }[/tex] < [tex]\frac{0.75-0.753}{0.004} } }[/tex] ) = P(Z < -0.75) = 1 - P(Z [tex]\leq[/tex] 0.75)
= 1 - 0.7734 = 0.2266
P(X [tex]\leq[/tex] 0.74) = P( [tex]\frac{X-\mu}{\sigma} } }[/tex] [tex]\leq[/tex] [tex]\frac{0.74-0.753}{0.004} } }[/tex] ) = P(Z [tex]\leq[/tex] -3.25) = 1 - P(Z < 3.25)
= 1 - 0.9994 = 0.0006
The above probability is calculated by looking at the value of x = 0.75 and x = 3.25 in the z table which has an area of 0.7734 and 0.9994 respectively.
Therefore, P(0.74 inch < X < 0.75 inch) = 0.2266 - 0.0006 = 0.226.
(c) Probability that a ball bearing is above the upper specification limit is given by = P(X > 0.76 inch)
P(X > 0.76) = P( [tex]\frac{X-\mu}{\sigma} } }[/tex] > [tex]\frac{0.76-0.753}{0.004} } }[/tex] ) = P(Z > -1.75) = 1 - P(Z [tex]\leq[/tex] 1.75)
= 1 - 0.95994 = 0.0401
The above probability is calculated by looking at the value of x = 1.75 in the z table which has an area of 0.95994.
(d) Probability that a ball bearing is below the lower specification limit is given by = P(X < 0.74 inch)
P(X < 0.74) = P( [tex]\frac{X-\mu}{\sigma} } }[/tex] < [tex]\frac{0.74-0.753}{0.004} } }[/tex] ) = P(Z < -3.25) = 1 - P(Z [tex]\leq[/tex] 3.25)
= 1 - 0.9994 = 0.0006
The above probability is calculated by looking at the value of x = 3.25 in the z table which has an area of 0.9994.
Using the normal distribution, it is found that there is a
a) 0.2734 = 27.34% probability that a ball bearing is between the target and the actual mean.
b) 0.226 = 22.6% probability that a ball bearing is between the lower specification limit and the target.
c) 0.0401 = 4.01% probability that a ball bearing is above the upper specification limit.
d) 0.0006 = 0.06% probability that a ball bearing is below the lower specification limit.
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.In this problem:
Mean of 0.753 inch, hence [tex]\mu = 0.753[/tex].Standard deviation of 0.004 inch, hence [tex]\sigma = 0.004[/tex]Item a:
This probability is the p-value of Z when X = 0.753 subtracted by the p-value of Z when X = 0.75.
X = 0.753:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{0.753 - 0.753}{0.004}[/tex]
[tex]Z = 0[/tex]
[tex]Z = 0[/tex] has a p-value of 0.5
X = 0.75:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{0.75 - 0.753}{0.004}[/tex]
[tex]Z = -0.75[/tex]
[tex]Z = -0.75[/tex] has a p-value of 0.2266
0.5 - 0.2266 = 0.2734
0.2734 = 27.34% probability that a ball bearing is between the target and the actual mean.
Item b:
This probability is the p-value of Z when X = 0.75 subtracted by the p-value of Z when X = 0.74, hence:
X = 0.75:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{0.75 - 0.753}{0.004}[/tex]
[tex]Z = -0.75[/tex]
[tex]Z = -0.75[/tex] has a p-value of 0.2266
X = 0.74:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{0.74 - 0.753}{0.004}[/tex]
[tex]Z = -3.25[/tex]
[tex]Z = -3.25[/tex] has a p-value of 0.0006.
0.2266 - 0.0006 = 0.226
0.226 = 22.6% probability that a ball bearing is between the lower specification limit and the target.
Item c:
This probability is 1 subtracted by the p-value of Z when X = 0.76, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{0.76 - 0.753}{0.004}[/tex]
[tex]Z = 1.75[/tex]
[tex]Z = 1.75[/tex] has a p-value of 0.9599.
1 - 0.9599 = 0.0401
0.0401 = 4.01% probability that a ball bearing is above the upper specification limit.
Item d:
This probability is the p-value of Z when X = 0.74, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{0.74 - 0.753}{0.004}[/tex]
[tex]Z = -3.25[/tex]
[tex]Z = -3.25[/tex] has a p-value of 0.0006.
0.0006 = 0.06% probability that a ball bearing is below the lower specification limit.
A similar problem is given at https://brainly.com/question/24663213
Video Example EXAMPLE 5 Where is the function f(x) = |x| differentiable? SOLUTION If x > 0, then |x| = and we can choose h small enough that x + h > 0 and hence |x + h| = . Therefore, for x > 0 we have
Answer:
Step-by-step explanation:
We say that a function is differentiable if it's derivative exists and it is continous over the domain of the original function. Recall that the function |x| is given as |x| = x if [tex]x\geq 0[/tex] and |x|=-x if x<0. In both cases, we have either the function x or -x, which are one-degree polynomials. So, we can find easily their derivatives. So if f(x) = |x|. Then, we have that
[tex]f'(x) = (x)' = 1 \text{if } x\geq 0[/tex]
[tex]f'(x) = (-x)' = -1 \text{if } x<0[/tex]
We should check what happens x=0, since that is the point where the definition of f changes. We can check that
[tex]\lim_{x\to 0^+} f'(x) = 1[/tex] and
[tex]\lim_{x\to 0^-} f'(x) = -1[/tex]
since this limits are not equal, the derivative is not continous at x=0. Hence, the derivative doesn't exist at x=0.
In this case, we say that |x| is differentiable at any x different from 0.
Find the missing side length. Show all of your work. I'll give you the brilliant crown if you answer the question correct.
Answer:
h = 6
Step-by-step explanation:
We need to use Pythagorean theorem.
h² + 8² = 10²
h² = 100 - 64 = 36
h = 6
The distance it takes a truck to stop can be
modeled by the function
Determine the value off, rounde
hundredth.
2
DONE
2.150 ?
d(u)=-
64.4f
d = stopping distance in feet
y = initial velocity in miles per hour
f= a constant related to friction
When the truck's initial velocity on dry pavement is
40 mph, its stopping distance is 138 ft.
Your question is not well presented (Refer below for correct question)
The distance it takes a truck to stop can be modeled by the function;
d(u) = 2.15u²/64.4f
Where
d = stopping distance in feet
u = initial velocity in miles per hour
f = a constant related to friction
When the truck's initial velocity on dry pavement is 40 mph, its stopping distance is 138 ft.
Determine the value of f.
Answer:
The friction constant is approximately 0.3871
Step-by-step explanation:
Given
Stopping distance, d(u) = 138ft
Initial Velocity, u = 40mph
Required
Friction constant, f.
To get the value of f, we need to simply substitute 40 for u and 138 for d(u) in the above expression.
In other words;
d(u) = 2.15u²/64.4f becomes
138 = 2.15 * 40²/64.4f
138 = 2.15 * 1600/64.4f
138 = 3440/64.4f
Multiply both sides by 64.4f
138 * 64.4f = 64.4f * 3440/64.4f
8887.2f = 3440
Divide both sides by 8887.2
8887.2f/8887.2 = 3440/8887.2
f = 3440/8887.2
f = 0.387073544
f = 0.3871 (Approximated)
Hence, the friction constant is approximately 0.3871
Answer:
first one is .39
second answer is C the last one
Step-by-step explanation:
Write an equation of the line that passes through (2,-1) and (-3,3)
Answer:
Y=-4/5x+3/5y
Step-by-step explanation: first get the gradient
3--1/-3-2
=4/-5
y-3/x+3=4/-5
-5y+15=4x+12
-5y=4x-3
y=4/-5x+3/5
if f(x)= 4x squared + 1 and g(x) = x squared -5 , find (f+g)(x)
Answer:
5x^2 -4
Step-by-step explanation:
f(x)= 4x^2+ 1
g(x) = x^2 - 5
(f+g)(x) = f(x) + g(x)
= 4x^2+ 1 + x^2 - 5
= 4x^2 + x^2 + 1- 5
= 5x^2 -4
Please answer this correctly
Answer:
The first one will be da answer
Step-by-step explanation:
happy to help you!
The distance between (-1,2) and (-2,1)
Answer:
About 10 i think
Step-by-step explanation:
idk
If f(x) = 4x^2 and g(x) = x+1, find (f.g)(x).
Answer:
4x^3 + 4x^2
Step-by-step explanation:
f(x) = 4x^2
g(x) = x + 1
=> (f.g)(x) = 4x^2*(x + 1) = 4x^3 + 4x^2
Hope this helps!
I have attached 3 different problems, please could you help. Thank you
Answer:
1a) 210
1b) 0.18
2) 17.7 will fit (> 17)
3) 515 cm^2
Step-by-step explanation:
1a) The sample of 25 is exactly 1/30 of the population of 750, so the number of orange cakes in the population will be 30 times the number in the sample.
The total number of orange cakes made on Monday is 30·7 = 210.
__
1b) The lowest number that will round to 5 is 4.5. The lowest probability that a cake is an orange cake is ...
4.5/25 = 18/100 = 0.18
The lower bound on the probability that the cake is orange is 0.18.
____
2) The working shown is partly correct.
The possible error in the box dimension is 1/2 cm, so the box may be as short as 28 -1/2 cm = 27.5 cm.
The possible error in the disc case dimension is 1/2 mm = 0.05 cm, so the case may be as thick as 1.5 +.05 = 1.55 cm.
The minimum number discs that will fit in the case is (27.5 cm)/(1.55 cm) = 17.7. Thus, 17 cases will definitely fit in the box.
____
3) The shaded area will be the largest when the outside rectangle is the largest and the inside rectangle is the smallest. Errors in dimensions can be as much as 1/2 cm, so the Outside rectangle can be 42.5 cm by 21.5 cm. The inside rectangle can be as small as 27.5 cm by 14.5 cm.
Then the shaded area can be as large as ...
(42.5)(21.5) -(27.5)(14.5) = 515 . . . . cm^2
