MATHEMATICS HIGH SCHOOL

A population has a mean of 300 and a standard deviation of 70. Suppose a sample of size 125 is selected and is used to estimate . Use z-table. What is the probability that the sample mean will be within +/- 8 of the population mean (to 4 decimals)? (Round z value in intermediate calculations to 2 decimal places.) What is the probability that the sample mean will be within +/- 17 of the population mean (to 4 decimals)? (Round z value in intermediate calculations to 2 decimal places.)

Answers

Answer 1

Answer:

Following are the calculation to the given points:

Using central limit theorem:

$\to \mu_{\bar{x}}= \mu = 300\n\n\to \sigma_{\bar{x}}=(\sigma)/(√(n))=(70)/(√(125))= 6.261\n\n$

For point a)

Within $\pm 8$ population mean: $\to P(\mu-8<\bar{x}<\mu+8)=P\left ( (\mu-8-\mu)/(6.261)Within [tex]\pm 17$

For point b)

Within $\pm 17$ population mean:

[tex]\to P(\mu-17<\bar{x}<\mu+17)=P\left ( \frac{\mu-17-\mu}{6.261}

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Answer 2

Answer:

Final answer:

To find the probability that the sample mean will be within a certain range, we need to calculate the z-scores and look them up in the z-table.

Explanation:

To find the probability that the sample mean will be within +/- 8 of the population mean, we need to convert this range to z-scores. The z-score formula is calculated by subtracting the population mean from the sample mean and dividing by the standard deviation divided by the square root of the sample size.

For +/- 8, the z-score will be (8 - 0) / (70 / sqrt(125)) = 2.09.

Using the z-table, we can find the probability associated with a z-score of 2.09, which is approximately 0.9828. Therefore, the probability that the sample mean will be within +/- 8 of the population mean is 0.9828.

Similarly, for +/- 17, the z-score will be (17 - 0) / (70 / sqrt(125)) = 4.18. Using the z-table, the probability associated with a z-score of 4.18 is nearly 1.0000. Hence, the probability that the sample mean will be within +/- 17 of the population mean is 1.0000.

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1. A rectangular picture 6cm by 8cm is enclosed by a frame 1/2cm wide. Calculate the area of the frame. A.15sq cm B. 20 sq cm C. 13 sq cm D. 16 sq cm E. 17

Answers

Answer:

The area of 1/2 cm wide frame = 48 - 35 = 13 cm²

Step-by-step explanation:

Length of rectangular picture frame = 6 cm

Width of rectangular picture frame = 8 cm

Area of rectangular picture frame = 6×8 cm² = 48 cm²

Length, excluding 1/2 cm wide frame of rectangular picture = (6–1/2-1/2) = 5 cm

Width, excluding 1/2 cm wide frame of rectangular picture = (8–1/2-1/2) = 7 cm

Area, excluding 1/2 cm wide frame of rectangular picture = 5×7 cm² = 35 cm²

Area of 1/2 cm wide frame = 48 - 35 = 13 cm²

Therefore, the area of 1/2 cm wide frame = 48 - 35 = 13 cm²

How do i solve this answer ugh please help

Answers

Separate the quadratic functions into two individual function in x which multiply together to equal the original quadratic.
Eg: x^2-3x-10=0 can be written as (x+2)(x-5)=0
The two solutions for x can then be determined by making each bracket equal to zero.
x+2=0 so x=-2x-5=0 so x=5

Find the value of 83 - [59 - (22 - 18)].
16
28
64

Answers

28 would be the correct answer to that equation

Answer:

Step-by-step explanation:

Using PEMDAS (order of operations) it would end up being 28

-x(7x - 8)

A. 6x^2 - 9x
B. -7x - 8x
C. -7x^2 + 8x
D. 7x + 8x

Answers

C. -7x ^ + 8x
the answer

Bryan is cooking cupcakes for a bake sale at his club. The recipe calls for 12 cups of flour. He has already put 3 cups of flour in the mixing bowl. How many more cups does he need to put in? Write an equation to represent this problem. Then solve.

Answers

Answer:

Bryan needs 9 cups more.

Step-by-step explanation:

Bryan is cooking cupcakes for a bake sale at his club. The recipe calls for 12 cups of flour.

Number of cups of flour, he already has = 3

Lets say he needs x cups more.

So, equation forms = $12=x+3$

=> $12-3=x$

x = 9

Hence, Bryan needs 9 cups more.

Answer:

9 cups

Step-by-step explanation:

12-3 = 9

The intensity of a radio signal from the radio station varies inversely as the square of the distance from the station. Suppose intensity is 8000 units at a distance of 2 miles. What will the intensity be at a distance of 11 miles? Round your answer to the nearest unit