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2 years ago

9

As sixth graders were entering Emerson middle school, there free supplies were being distributed. Every 5the sixth grader that e

ntered the building recived a free pencil. Every 15th sixth grader that entered the building received a free note book. Every 50th sixth grader that entered the building recived a free t-shrit. if there were 260 sixth graders entering the building, how many recived a free pencil, how many received a free notebook, how many students received all three supplies for free?

1 answer:

jok3333 [9.3K]2 years ago

4 0

52 received a pencil
17 received a notebook
5 received a t shirt

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Where are the asymptotes for the following function located? f(x)= 7/ x^2-2x-24

Zarrin [17]

Factor the demoniator:-

x^2 -2x - 24 = (x - 6)(x + 4)

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The access code for a car's security system consists of sixsix digits. the first digit cannot be zerozero and the last digit mu

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3 years ago

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

Well I need help please

marissa [1.9K]

change to improper fractions

77 1/2 = 145/2

2 1/4 = 9/4

divide

145/2 divide 9/4

copy dot flip

145/2 * 4/9

580/18

290/9

32 2/9

it will fit into 32 bags with 2/9 left over

We have 32 full bags * 2.25 = $72.00

8 0

3 years ago

Can someone help me with this please?

natima [27]

150/250! Hope this helps :D

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3 years ago