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3 years ago

The expression 1.7j-3.4 factored is?

Mathematics

1 answer:

marshall27 [118]3 years ago

3 0

1.7(j-2)

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Tess created a Pythagorean triple of (24, 32, 40) by multiplying the known Pythagorean triple 3, 4, and 5 by 8. Is Tess correct?

pentagon [3]

Answer:

(24, 32, 40) is a Pythagorean triplet.

Option A) Yes, as long as each number in the Pythagorean triple is multiplied by the same whole number.

Step-by-step explanation:

We are given the following in the question:

Pythagorean triplet: (3,4,5)

New Pythagorean triplet obtained = (24, 32, 40)

Pythagoras theorem:

For a right angles triangle the sum of square of sides of triangle is equal to the square of hypotenuse of triangle.

Verification:

$24^2 + 32^2\\=576 + 1024\\=1600\\=40^2$

Yes, Tessa is correct.

(24, 32, 40) is a Pythagorean triplet.

Pythagorean triple be created using multiplies of a known Pythagorean triple

Option A) Yes, as long as each number in the Pythagorean triple is multiplied by the same whole number.

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3 years ago

Read 2 more answers

20,612,439 in scientific notation

Evgen [1.6K]

2.0612439*10^7 should be your answer

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3 years ago

A toy store wants to give away free goody bags to their customers. They have 60 lollipops and 72 stickers to give away. They wan

Flura [38]

60 because each bag has to be the same

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2 years ago

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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2 years ago

Solve the following system of linear equations:<br> 4x - 2y = 3

Feliz [49]

Answer:

follow me for a follow back!

Step-by-step explanation:

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3 years ago