HELPP QUICK PLS <br> ILL GIVE BRAINLIEST

A greeting card company can produce a box of cards for $7.50. If the initial investment by the company was $50,000, how many box

larisa86 [58]

Answer:

If cost of one box of card is $7.50 then the no of cards produced is 6667 cards and if cost of one box of card is $10.50 then the no of cards produced is 4762 cards

Step-by-step explanation:

The cost of one box of cards = $7.50

Initial investment = $50,000

No of box of cards produced = Initial investment / cost of one box of card

No of box of cards produced = 50,000/7.50

No of box of cards produced = 6667 cards.

if the cost of one box of cards is increased = $10.50

No of box of cards produced = Initial investment / cost of one box of card

No of box of cards produced = 50,000/10.50

No of box of cards produced = 4762 cards.

So, if cost of one box of card is $7.50 then the no of cards produced is 6667 cards and if cost of one box of card is $10.50 then the no of cards produced is 4762 cards

5 0

3 years ago

The football team ordered 2 pizzas, they didn't eat 1/12 of one pizza and 2/4 of the other. How much pizza was left?

kodGreya [7K]

To equally divide the pizzas, let's use a similar point to base on the remaining pizzas. On the first pizza 1/12 was not eaten an the other was 2/4. Use 12 as the base parts. 1/12 +2/4. 2/4 is equal to 1/2, so let's used 6/12 (also 1/2) to replace 2/4 to add both fractions. 1/12+6/12=7/12.

3 0

3 years ago

Assume, for the sake of this question, that the data were collected through a well-designed, well-implemented random sampling me

amid [387]

Answer:

Since the pvalue of the test is 0.2743 > 0.1, the threshold probably was met.

Step-by-step explanation:

The widget manufacturing company had established a threshold of 60% preferring the proposed new widget to move forward with producing the new widgets.

This means that at the null hypothesis we test if the proportion is at least 60%, that is:

$H_{0}: p \geq 0.6$

And the alternate hypothesis is:

$H_{a}: p < 0.6$

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

0.6 is tested at the null hypothesis:

This means that:

$\mu = 0.6$

$\sigma = \sqrt{0.6*0.4}$

Three hundred thirty-eight of 575 respondents reported preferring the proposed new widget.

This means that $n = 575, X = \frac{338}{575} = 0.5878$

Value of the test-statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{0.5878 - 0.6}{\frac{\sqrt{0.6*0.4}}{\sqrt{575}}}$

$z = -0.6$

Pvalue of the test and decision:

We want to find the probability of a proportion of 0.5878 or lower, which is the pvalue of z = -0.6.

Looking at the z-table, z = -0.6 has a pvalue of 0.2743.

Since 0.2743 > 0.1, the threshold probably was met.

3 0

3 years ago

How to answer a problem that says 5 ro the 4 power over 5

Eddi Din [679]

$\bf ~~~~~~~~~~~~\textit{negative exponents} \\\\ a^{-n} \implies \cfrac{1}{a^n} \qquad \qquad \cfrac{1}{a^n}\implies a^{-n} \qquad \qquad a^n\implies \cfrac{1}{a^{-n}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \cfrac{5^4}{5}\implies 5^4\cdot 5^{-1}\implies 5^{4-1}\implies 5^3\implies 125$

4 0

3 years ago

Charlie is at a small air field watching for the approach of a small plane with engine trouble. He sees the plane at an angle of

lubasha [3.4K]

(32°) = 1700 feet / x => x = 1700 feet / tan(32°) = 1700 feet / 0.625 = 2720 feet.

6 0

3 years ago

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