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3 years ago

Isabella is ordering invitations

Mathematics

1 answer:

MrMuchimi3 years ago

6 0

Clear Print: y = 2x
Sue: y = 36 + 0.50x

Clear Print: y = 2(60) = $120
Sue: y = 36 + 0.50(60) = $66

Sue offers the better price for 60 invitations

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During a football game a concession stand sold a family three hamburgers and two hotdogs for a total of $13 it's sold another fa

guajiro [1.7K]

Answer:

The price of each hamburger is $3

The price of each hot dogs is $2 .

Step-by-step explanation:

Given as :

The total price of 3 hamburger and 2 hot dogs = $13

The total price of 2 hamburger and 5 hot dogs = $16

Let The price of each hamburger = $x

Let The price of each hot dogs = $y

<u>Now, According to question</u>

3 x + 2 y = 13 .........A

2 x + 5 y = 16 .......B

Now, Solving to eq A and B

3 × (2 x + 5 y ) - 2 × (3 x + 2 y ) = 3 × 16 - 2 × 13

Or, (6 x + 15 y) - (6 x + 4 y) = 48 - 26

Or, (6 x - 6 x) + (15 y - 4 y) = 22

Or, 0 + 11 y = 22

∴ y = $\dfrac{22}{11}$

i.e y = $2

so, The price of each hot dogs = y = $2

<u>Now, Put the value of y into eq B</u>

i.e 2 x + 5 y = 16

or, 2 x + 5 × 2 = 16

or, 2 x = 16 - 10

or, 2 x = 6

∴ x = $\dfrac{6}{2}$

i.e x = $3

So, The price of each hamburger = x = $3

Hence, The price of each hamburger is $3 and The price of each hot dogs is $2 . Answer

3 0

4 years ago

Helpp noww pleasee!! 30 pointss

masha68 [24]

Answer:

729

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Solve for t.

finlep [7]

Your answer would be A)

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7 0

3 years ago

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At 95% confidence, how large a sample should be taken to obtain a margin of error of 0.03 for the estimation of a population pro

Gnom [1K]

Answer:

A sample of 1068 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 $1 - \frac{\alpha}{2}$ .

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 $Z = 1.96$ .

At 95% confidence, how large a sample should be taken to obtain a margin of error of 0.03 for the estimation of a population proportion?

We need a sample of n.

n is found when M = 0.03.

We have no prior estimate of $\pi$ , so we use the worst case scenario, which is $\pi = 0.5$

Then

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

$0.03 = 1.96\sqrt{\frac{0.5*0.5}{n}}$

$0.03\sqrt{n} = 1.96*0.5$

$\sqrt{n} = \frac{1.96*0.5}{0.03}$

$(\sqrt{n})^{2} = (\frac{1.96*0.5}{0.03})^{2}$

$n = 1067.11$

Rounding up

A sample of 1068 is needed.

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3 years ago

If,

lora16 [44]

Answer

What is the question?

Explanation

Thanks for the free points

Have a great day/night ❤

6 0

3 years ago