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maxonik [38]
3 years ago
6

Factorise-              6y-15y^2     HELP

Mathematics
2 answers:
Citrus2011 [14]3 years ago
7 0
We will take the common terms in the expressions-

Here in the expression, 3y is a common term.
So,

6y - 15y^{2}

3y(2 - 5y)

So, the answer is 3y(2 -5y)
Ulleksa [173]3 years ago
7 0
6y-15y^2
3y(2-5y)                                                          (taking 3y as common)
clearly 3y(2-5y) cannot be factorised further hence this is the answer
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The surface areas of the two solids shown above are equal.<br> O A. True<br> O B. False
nikdorinn [45]

Answer:

False

Step-by-step explanation:

If we were to flip the solid on the left so its parallel to the solid on the right, we would be able to compare the two more easier.

We can see that the right solid has dimensions of:

L = 1 cm

W = 3 cm

H = 5 cm

The left solid has dimensions of:

L = 1

W = 2

H = 7

If we were to add these all up, they would not equal.

R: 1 + 3 + 5 = 9

L: 1 + 2 + 7 = 10

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The answer is 8/17 because you have to change the fractions to improper fractions then change the division sign to a multiplication sign. Then change the second fraction by flipping the two numbers then multiply and reduce.
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Nadya [2.5K]

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Step-by-step explanation:


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3 years ago
There are big spenders among University of Alabama football season ticket holders. This data set Roll Tide!! shows the dollar am
Bas_tet [7]

Using the z-distribution, as we have a proportion, the 95% confidence interval is (0.2316, 0.3112).

<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.

We also consider that 130 out of the 479 season ticket holders spent $1000 or more at the previous two home football games, hence:

n = 479, \pi = \frac{130}{479} = 0.2714

Hence the bounds of the interval are found as follows:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2714 - 1.96\sqrt{\frac{0.2714(0.7286)}{479}} = 0.2316

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2714 + 1.96\sqrt{\frac{0.2714(0.7286)}{479}} = 0.3112

The 95% confidence interval is (0.2316, 0.3112).

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

7 0
2 years ago
Please help!!! will mark brainliest
faust18 [17]

Your answer is 7.

I subtracted 5 from 12 to find the missing number for KL that is 7.

Therefore, your answer is 7

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