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3 years ago

What is the prime factorization of 325

Mathematics

1 answer:

Tasya [4]3 years ago

8 0

The Prime Factorization of 325
Answer : 5^2 • 13
hope this helps!!!

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7 + 9y < -3 (y - 5) + 4

Anuta_ua [19.1K]

Answer:

y<1

Step-by-step explanation:

12y/12

12/12

5 0

3 years ago

The data given to the right includes data from candies, and of them are red. The company that makes the candy claims that % of

Dafna1 [17]

Using the z-distribution, the 95% confidence interval for the percentage of red candies is of (7.84%, 33.18%). Since 33% is part of the interval, there is not enough evidence to conclude that the claim is wrong.

<h3>What is a confidence interval of proportions?</h3>

A confidence interval of proportions is given by:

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

In which:

$\pi$ is the sample proportion.

z is the critical value.

n is the sample size.

In this problem, we have a 95% confidence level, hence $\alpha = 0.95$ , z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$ , so the critical value is z = 1.96.

Researching this problem on the internet, 8 out of 39 candies are red, hence the sample size and the estimate are given by:

$n = 39, \pi = \frac{8}{39} = 0.2051$

Hence the bounds of the interval are:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2051 - 1.96\sqrt{\frac{0.2051(0.7949)}{39}} = 0.0784$

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2051 + 1.96\sqrt{\frac{0.2051(0.7949)}{39}} = 0.3318$

As a percentage, the 95% confidence interval for the percentage of red candies is of (7.84%, 33.18%). Since 33% is part of the interval, there is not enough evidence to conclude that the claim is wrong.

More can be learned about the z-distribution at brainly.com/question/25890103

#SPJ1

7 0

2 years ago

What number is 80% of 90?

Lerok [7]

The answer is 72

b/c 80% is equal to .80

so you do 90x.80

4 0

3 years ago

Let z=3+i, then find a. Z² b. |Z| c.<img src="https://tex.z-dn.net/?f=%5Csqrt%7BZ%7D" id="TexFormula1" title="\sq

zysi [14]

Given z = 3 + i, right away we can find

(a) square

z ² = (3 + i )² = 3² + 6i + i ² = 9 + 6i - 1 = 8 + 6i

(b) modulus

|z| = √(3² + 1²) = √(9 + 1) = √10

(d) polar form

First find the argument:

arg(z) = arctan(1/3)

Then

z = |z| exp(i arg(z))

z = √10 exp(i arctan(1/3))

z = √10 (cos(arctan(1/3)) + i sin(arctan(1/3))

Any complex number has 2 square roots. Using the polar form from part (d), we have

√z = √(√10) exp(i arctan(1/3) / 2)

and

√z = √(√10) exp(i (arctan(1/3) + 2π) / 2)

Then in standard rectangular form, we have

$\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right)\right)$

and

$\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right)\right)$

We can simplify this further. We know that z lies in the first quadrant, so

0 < arg(z) = arctan(1/3) < π/2

which means

0 < 1/2 arctan(1/3) < π/4

Then both cos(1/2 arctan(1/3)) and sin(1/2 arctan(1/3)) are positive. Using the half-angle identity, we then have

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

and since cos(x + π) = -cos(x) and sin(x + π) = -sin(x),

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

Now, arctan(1/3) is an angle y such that tan(y) = 1/3. In a right triangle satisfying this relation, we would see that cos(y) = 3/√10 and sin(y) = 1/√10. Then

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10+3\sqrt{10}}{20}}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10-3\sqrt{10}}{20}}$

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}$

So the two square roots of z are

$\boxed{\sqrt z = \sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}$

and

$\boxed{\sqrt z = -\sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}$

3 0

3 years ago

Read 2 more answers

Evaluate the Expression When X=2, 3x-1=

Alex777 [14]

Answer:

Step-by-step explanation:

If X equals 2, then 3 times X would be 6. This is because if a number and a variable are put together, you have to multiply them. 6-1 is 5.

4 0

3 years ago