HELP ME PLS ILL MARK AS BRAINLESS

Mathematics

2 answers:

Arada [10]2 years ago

6 0

Answer:

Step-by-step explanation:

Ratios are normally part to part which knocks off answer choices C and D.

Ratios are written with the smaller value on the left and the large value on the right so it can't be B

So it has to be A

faltersainse [42]2 years ago

3 0

Answer:

A: 12:18

Step-by-step explanation:

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POSSIBLE POINTS: 11.76

natta225 [31]

Answer:

B and D.

Brainliest please. :)

3 0

2 years ago

Timed test

marysya [2.9K]

Answer:

The correct answer should be 5,130

Step-by-step explanation:

the 3 in 513 is in the ones place, if it were ten times greater than 3 that would be 30, or a 3 in the tens place, the 3 in the number 5,130 is in the tens place, making that the correct choice

4 0

2 years ago

tyreee bought a collection comic book for 49.62 last year.this year he would sold it for 52.10 find the percent of change.

wlad13 [49]

Answer:

The percentage increased by approximately 5%.

Step-by-step explanation:

52.10 - 49.62 = 2.48

2.48/49.62 x 100 = 4.998%

The percentage increase in the price is approximately 5%. Hope this helps! :)

3 0

2 years ago

In a randomly selected sample of 100 students at a University, 81 of them had access to a computer at home. Give the value of th

astra-53 [7]

Answer:

The value of the standard error for the point estimate is of 0.0392.

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$