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3 years ago

Is 13/10 irrational or rational

Mathematics

2 answers:

Akimi4 [234]3 years ago

5 0

Answer:

Yes

Step-by-step explanation:

Both 13 and 10 are integers, meaning 13.10 is rational.

motikmotik3 years ago

4 0

It is rational 13/10

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Combine the following sets of terms.

DanielleElmas [232]

Let's simplify step-by-step To understand :)

3x2−7x−(x2+3x−9)

Distribute the Negative Sign:

=3x2−7x+−1(x2+3x−9)

=3x2+−7x+−1x2+−1(3x)+(−1)(−9)

=3x2+−7x+−x2+−3x+9

Combine Like Terms:)

=3x2+−7x+−x2+−3x+9

=(3x2+−x2)+(−7x+−3x)+(9)

=2x2+−10x+9

Answer :)

=2x2−10x+9

HOPE THIS HELPS

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3 years ago

2 friends go out to lunch and order 1 large pizza.The friends divided the pizza evenly.Find how much pizza each friend got as a

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Does It Say How Much Slices There Are ?

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3 years ago

Read 2 more answers

64 percent of what number is 48

sergey [27]

The answer I got was 75%

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3 years ago

Find the sum of partial fraction for x^4/(x-1)(x²-4)

Aneli [31]

Answer:

see attachment

Step-by-step explanation:

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3 years ago

There are 2,000 eligible voters in a precinct. A total of 500 voters are randomly selected and asked whether they plan to vote f

Ann [662]

Answer:

$0.7 - 2.58 \sqrt{\frac{0.7(1-0.7)}{500}}=0.647$

$0.7 + 2.58 \sqrt{\frac{0.7(1-0.7)}{500}}=0.753$

And the 99% confidence interval would be given (0.647;0.753).

So the correct answer would be:

a. 0.647 and 0.753

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Solution to the problem

The estimated population proportion for this case is:

$\hat p = \frac{350}{500}=0.7$

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.58$

And replacing into the confidence interval formula we got:

$0.7 - 2.58 \sqrt{\frac{0.7(1-0.7)}{500}}=0.647$

$0.7 + 2.58 \sqrt{\frac{0.7(1-0.7)}{500}}=0.753$

And the 99% confidence interval would be given (0.647;0.753).

So the correct answer would be:

a. 0.647 and 0.753

7 0

3 years ago