MATHEMATICS COLLEGE

Problem 9-10 The elongation of a steel bar under a particular tensile load may be assumed to be normally distributed, with a mean of .06 in. and standard deviation of .008 in. A sample of n=100 bars is subjected to the test. Find the probability that the sample mean elongation is between .0585 in. and .0605 in.

Answers

Answer 1

Answer:

Answer:

The answer is 0.7036.

Step-by-step explanation:

Check the attached file for the computations.

Answer 2

Answer:

The probability that the mean life of a random sample mean elongation is between .0585 in. and .0605 in. is 70.57%

What is z score?

Z score is used to determine by how many standard deviations the raw score is above or below the mean.

It is given by:

z = (raw score - mean) / (standard deviation÷√sample)

Mean = 0.06, standard deviation = 0.008, sample = 100.

For x = 0.0585:

z = (0.0585 - 0.06)/ (0.008 ÷√100) = -1.88

For x = 7.2:

z = (0.0605 - 0.06)/ (0.008 ÷√100) = 0.623

P(-1.88 < z < 0.63) = P(z < 0.63) - P(z < -1.88) = 0.7357 - 0.03 = 0.7057

The probability that the mean life of a random sample mean elongation is between .0585 in. and .0605 in. is 70.57%

Find out more on z score at: brainly.com/question/25638875

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Bruce earns 14.45/h and works 40 hours a week. What is his gross annual income

Answers

Answer:

$30056

Step-by-step explanation:

First, you have to do $14.45 times 40 to get how much he makes in a week, and you get $578. Then do 578 x 52 as there are 52 weeks in a year. Then you would get the answer $30056 as his yearly income.

Answer:

$30,056.00

Step-by-step explanation:

Took the quiz

Find the slope of the line y=1

Answers

The Slope is 0. The graph will not rise or fall

True it will go across the positive 1 line on the y axis

About 4 out of every 5 people are right-handed. If there are 30 students in a class, how many would you expect to be right-handed? How many to be left-handed?

Answers

Answer:

You would expect 24 right handed and 6 left handed

Step-by-step explanation:]

5 x 6 =30

A survey of the mean number of cents off that coupons give was conducted by randomly surveying one coupon per page from the coupon sections of a recent San Jose Mercury News. The following data were collected: 20¢; 75¢; 50¢; 65¢; 30¢; 55¢; 40¢; 40¢; 30¢; 55¢; $1.50; 40¢; 65¢; 40¢. Assume the underlying distribution is approximately normal. (a) Determine the sample mean in cents (Round to 3 decimal places)

Answers

Answer:

53¢

Step-by-step explanation:

First, I'll put these in order.

20¢;30¢; 30¢;75¢;40¢;40¢;40¢;40¢;50¢;55¢55¢65¢;65¢; $1.50;

Then, I'll combine like terms.

30+30=60

40+40+40+40(or 40 x 4)=160

55+55=110

65+65=130

60+160+110+130+20+75+50+$1.50=$7.55/14=53¢

PLZ correct me if i'm wrong :-D

Pleaseeeee helpppppppppppp

Answers

9514 1404 393

Answer:

maximum height: 26.5 ft
air time: 2.5 seconds

Step-by-step explanation:

I find the easiest way to answer these questions is to use a graphing calculator. It can show you the extreme values and the intercepts. The graph below shows the maximum height is 26.5 ft. The time in air is about 2.5 seconds.

If you prefer to solve this algebraically, you can use the equation of the axis of symmetry to find the time of the maximum height:

t = -b/(2a) = -(40)/(2×-16) = 5/4

Then the maximum height is ...

h(5/4) = -16(5/4)² +40(5/4) +1.5 = -25 +50 +1.5 = 26.5 . . . feet

Now that we know the vertex of the function, we can write it in vertex form:

h(t) = -16(t -5/4)² +26.5

Solving for the value of t that makes this zero, we get ...

0 = -16(t -5/4)² +26.5

16(t -5/4)² = 26.5

(t -5/4)² = 26.5/16 = 1.65625

Then ...

t = 1.25 +√1.65625 ≈ 2.536954