5th grade math. Correct answer will be marked brainliest.

Mathematics

1 answer:

vitfil [10]3 years ago

4 0

Answer: The answer would be (0,3)

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What is the greatest common factor of 42a^5b^3, 35a^3b^4, and 42ab^4? 7ab^3 6a^4b 42a^5b^4 77a^8b^7

Blababa [14]

$42a^5b^3=7ab^3\cdot6a^4\\\\35a^3b^4=7ab^3\cdot5a^2b\\\\42ab^4=7ab^3\cdot6b\\\\\boxed{GCF(42a^5b^3,\ 35a^3b^4,\ 42ab^4)=7ab^3}$

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3 years ago

Read 2 more answers

How many of the positive divisors of 2160 are multiples of 3?

TEA [102]

30 of the divisors are multiples of 3

The factors of 2160 are 1, 2160, 2, 1080, 3, 720, 4, 540, 5, 432, 6, 360, 8, 270, 9, 240, 10, 216, 12, 180, 15, 154, 16, 135, 18, 120, 20, 108, 24, 90, 27, 80, 30, 72, 36, 60, 40, 54, 45, 48
These are all divisors of 2160 but only 30 are multiples of 3.
The ones that are NOT are 1, 2, 4, 5, 8, 10, 16, 20, 40 and 80

You can also do this using prime factors 2x2x2x2x3x3x3x5 = 2160 but it is harder to explain.

5 0

3 years ago

Use the substitution method to determine the solution type of the following system.

Elis [28]

The answer to this is C I think, I'm usually always correct so trust me k?

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3 years ago

The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e

irakobra [83]

Answer:

We need a sample size of at least 383.

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:

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

85% confidence level

So $\alpha = 0.15$ , z is the value of Z that has a pvalue of $1 - \frac{0.15}{2} = 0.925$ , so $Z = 1.44$ .

How large a sample would be required in order to estimate the fraction of tenth graders reading at or below the eighth grade level at the 85% confidence level with an error of at most 0.03

We need a sample size of at least n.

n is found with $M = 0.03, \pi = 0.21$