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2 years ago

9

Elaine found the following in her pocket. How much money was in her pocket

1 answer:

jenyasd209 [6]2 years ago

3 0

Answer:

we need the numbers... or like the coins, or something to solve it with. you need to include what was actually in here pocket lol

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In one year a dog’s weight increased from 64 to 80. By what percentage did the dogs weight increase?

kicyunya [14]

80 divided by 64 then add % to ur answer
So then you will have your percentage to the dogs weight increase

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2 years ago

Question 2 Please help. Find the arc length for the following:

harkovskaia [24]

9.4 would be your answer

4 0

2 years ago

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A high school sports team is ordering sports drinks online from a company. The price of sports drinks varies, based on the numbe

Elena-2011 [213]

The amount of drinks is a linear function of the number of drinks. The maximum amount that can be purchased is 100 soft drinks.

Let

$y \to$ amount of drinks

$x \to$ number of drinks

For drinks not more than 50

$y = 10 + 0.8x$

For drinks more than 50.

$y = 15 + 0.7x$

For a purchase of $85, it means that:

$y = 85$

So, we solve for x in both equations.

$y = 10 + 0.8x$

$85 = 10 + 0.8x$

Collect like terms

$0.8x = 85-10$

$0.8x = 75$

Divide through by 0.8

$x = 93.75$

$x = 94$ --- approximated

$y = 15 + 0.7x$

$85 =15 + 0.7x$

Collect like terms

$0.7x = 85 - 15$

$0.7x = 70$

Divide by 0.7

$x = 100$

By comparing both values:

$x = 100$

$x = 94$

$100 > 94$

Hence, the maximum amount that can be purchased is 100

Read more about functions at:

brainly.com/question/20286983

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3 years ago

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Which statement describes the graph of f(x) = [x] - 2 on [0, 3)?

Dima020 [189]

Answer:

A

Step-by-step explanation:

im on edg.

4 0

3 years ago

Skipping Lunch A nutritionist wishes to determine, within 3%, the true proportion of adults who do not eat any lunch. If he wish

mixas84 [53]

Answer:

A sample of at least 545 adults is needed.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

$\pi = 0.15$

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

If he wishes to be 95% confident that his estimate contains the population proportion, how large a sample will be necessary?

We need a sample of at least n.

n is found when M = 0.03. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.03 = 1.96\sqrt{\frac{0.15*0.85}{n}}$

$0.03\sqrt{n} = 1.96\sqrt{0.15*0.85}$

$\sqrt{n} = \frac{1.96\sqrt{0.15*0.85}}{0.03}$

$(\sqrt{n})^{2} = (\frac{1.96\sqrt{0.15*0.85}}{0.03})^{2}$

$n = 544.23$

Rounding up

A sample of at least 545 adults is needed.

6 0

2 years ago