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3 years ago

14

Im needing help with this math question

1 answer:

Zepler [3.9K]3 years ago

5 0

Answer:

4 weeks = 105

16 weeks = 42

24 weeks = 0

Step-by-step explanation:

the function is missing the 'w'

it should be : C(w) = 126 - 5.25w

'w' is the number of weeks

Substitute number of weeks in the 'w' spot

first one is 4 weeks, so

C(w) = 126 - 5.25(4)

= 126 - 21

= 105

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Answer:

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Step-by-step explanation:

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3 years ago

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3 years ago

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A garden is in the shape of a rectangle and has a perimeter of 58 yards. The length of the garden is 5 more than twice the width

slega [8]

$P=2l+2w$
$l=2w+5$
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Check
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3 years ago

What is equivalent to 3^4/3^7

Misha Larkins [42]

i believe your answer will be 0.04 or 4/100

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3 years ago

The manager of a pizza chain in Albuquerque, New Mexico, wants to determine the average size of their advertised 20-inch pizzas.

7nadin3 [17]

Answer:

The sample must be of at least 48 pizzas.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.9}{2} = 0.05$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.05 = 0.95$ , so $z = 1.645$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

How large a sample must she take if she wants the margin of error to be under 0.5 inch?

She needs a sample of at least n, in which is found when $M = 0.5, \sigma = 2.1$

$M = z*\frac{\sigma}{\sqrt{n}}$

$0.5 = 1.645*\frac{2.1}{\sqrt{n}}$

$0.5\sqrt{n} = 2.1*1.645$

$\sqrt{n} = \frac{2.1*1.645}{0.5}$

$(\sqrt{n})^{2} = (\frac{2.1*1.645}{0.5})^{2}$

$n = 47.73$

Rounding up

The sample must be of at least 48 pizzas.

3 0

3 years ago