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3 years ago

How am I supposed to do that??

Mathematics

1 answer:

marissa [1.9K]3 years ago

5 0

Answer:

not a solution

Step-by-step explanation:

y = -3/4 x +3

Is (3,2) a solution?

Substitute the point into the equation and see if it is true

2 = -3/4(3) +3

2 = -9/4 +3

2 = -9/4 +16/4

2 = 7/4

False

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Use the image to answer the question below. Which equation represents circle A?

Maksim231197 [3]

Answer is in the file below

tinyurl.com/wpazsebu

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3 years ago

A researcher wants to estimate the percentage of all adults that have used the Internet to seek pre-purchase information in the

lara31 [8.8K]

Answer:

The sample size is 1875.

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:

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

Sampling error of 0.03.

This means that $M = 0.03$

99.74% confidence level

So $\alpha = 0.0026$ , z is the value of Z that has a pvalue of $1 - \frac{0.0026}{2} = 0.9987$ , so $Z = 3$ .

25% of all adults had used the Internet for such a purpose

This means that $\pi = 0.25$

What is the sample size

The sample size is n. So

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

$0.03 = 3\sqrt{\frac{0.25*0.75}{n}}$

$0.03\sqrt{n} = 3\sqrt{0.25*0.75}$

Simplifying by 0.03 both sides

$\sqrt{n} = 100\sqrt{0.25*0.75}$

$(\sqrt{n})^2 = (100\sqrt{0.25*0.75})^2$

$n = 1875$

The sample size is 1875.

7 0

3 years ago

Someone help me please

MariettaO [177]

Answer:

5000÷yegey÷_×5/2

= 20000023

3 0

3 years ago

Decrease 845g by 62%

Phoenix [80]

845* (100-62)%
=845*38%
=845*38/100
=32110/100=321,1

4 0

3 years ago

Read 2 more answers

Is x greater than,less than,or equal to 94?

vovangra [49]

The answer is x=94...

Explanation:

Because x and 94 are vertically opposite angles..As we know vertically opposite angles are always equal.

Hope it will help u...

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3 years ago

Read 2 more answers