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schepotkina [342]

3 years ago

5

A box has dimensions of 14 inches long, 1.5 feet wide, and

1 answer:

Liula [17]3 years ago

7 0

1764 inches cubed. First you multiply 1.5 times 12 so that it is in inches. Then u get 18. U multiply 14 by 18 by 7 and u get 1764.

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Miss Guydan and her friend Alyssa are collecting food for the local food bank. Their goal is to collect 2100 pounds of food. The

Serjik [45]

Answer: Required equation: 400+25x = 2100

Variable: x = Number of days to donate 2100 pounds.

Step-by-step explanation:

Let x = Number of days.

y = Total pounds of food donated in x days.

Given: Initial amount of food = 400 pounds

Pounds of food donated per day = 25

Total food = Initial amount of food + (Pounds of food donated per day ) (Number of days)

y = 400+25x

Their goal is to collect 2100 pounds.

2100 = 400+25x [Substitute y= 2100]

Required equation: 400+25x = 2100

Variable: x = Number of days.

4 0

2 years ago

Kevin has $24 to buy a gift for his cousin. He found a gift for $22. With 5% sales tax added on, will Kevin have enough money to

mojhsa [17]

Answer:

Yes; $23.10

Step-by-step explanation:

5% of 22 is 1.10 and if you add the tax to the give price it is less than $24

8 0

2 years ago

Change to a fraction 3.025

scoray [572]

Answer:

3.025 as a fraction is 121/40

3 0

3 years ago

Read my name to understand this?

Molodets [167]

Answer:

Hello :) 12345678910

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

A simple random sample of items resulted in a sample mean of . The population standard deviation is . a. Compute the confidence

Varvara68 [4.7K]

Answer:

(a): The 95% confidence interval is (46.4, 53.6)

(b): The 95% confidence interval is (47.9, 52.1)

(c): Larger sample gives a smaller margin of error.

Step-by-step explanation:

Given

$n = 30$ -- sample size

$\bar x = 50$ -- sample mean

$\sigma = 10$ --- sample standard deviation

Solving (a): The confidence interval of the population mean

Calculate the standard error

$\sigma_x = \frac{\sigma}{\sqrt n}$

$\sigma_x = \frac{10}{\sqrt {30}}$

$\sigma_x = \frac{10}{5.478}$

$\sigma_x = 1.825$

The 95% confidence interval for the z value is:

$z = 1.960$

Calculate margin of error (E)

$E = z * \sigma_x$

$E = 1.960 * 1.825$

$E = 3.577$

The confidence bound is:

$Lower = \bar x - E$

$Lower = 50 - 3.577$

$Lower = 46.423$

$Lower = 46.4$ --- approximated

$Upper = \bar x + E$

$Upper = 50 + 3.577$

$Upper = 53.577$

$Upper = 53.6$ --- approximated

<em>So, the 95% confidence interval is (46.4, 53.6)</em>

Solving (b): The confidence interval of the population mean if mean = 90

First, calculate the standard error of the mean

$\sigma_x = \frac{\sigma}{\sqrt n}$

$\sigma_x = \frac{10}{\sqrt {90}}$

$\sigma_x = \frac{10}{9.49}$

$\sigma_x = 1.054$

The 95% confidence interval for the z value is:

$z = 1.960$

Calculate margin of error (E)

$E = z * \sigma_x$

$E = 1.960 * 1.054$

$E = 2.06584$

The confidence bound is:

$Lower = \bar x - E$

$Lower = 50 - 2.06584$

$Lower = 47.93416$

$Lower = 47.9$ --- approximated

$Upper = \bar x + E$

$Upper = 50 + 2.06584$

$Upper = 52.06584$

$Upper = 52.1$ --- approximated

<em>So, the 95% confidence interval is (47.9, 52.1)</em>

Solving (c): Effect of larger sample size on margin of error

In (a), we have:

$n = 30$ $E = 3.577$

In (b), we have:

$n = 90$ $E = 2.06584$

<em>Notice that the margin of error decreases when the sample size increases.</em>

4 0

3 years ago