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Makovka662 [10]
2 years ago
8

What is the least common multiple of 10 and 27

Mathematics
1 answer:
lara [203]2 years ago
7 0

Answer:

270 ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ

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Please help me!!!
Alika [10]

Answer:

Her GPA is 3

Step-by-step explanation:

an A is 4 points

B is 3 points

C is 2 points

D is 1 point

She has A, A, A, B, C, D for her grades. (6 grades total)

That equals

4 + 4 + 4 + 3 + 2 + 1 = 18 points

Then we divide the points by the number of grades we're looking at.

18/6 = 3

Her grade point average is 3.

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Find the equation of the line: parallel to 3x - y = 11 through (-2,0)
aleksandr82 [10.1K]

Answer:

y=3x+6

Step-by-step explanation:

The slope of the line is 3 and the equation will be y=3x+6

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What is the LCM of two numbers that have no common factors greater than 1
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If two numbers have no common factor greater then one, then their LCM is the two numbers multiplied together. Example: 9 and 14 have no common factors. Their LCM <span>is 9 x 14, which is 126.</span>
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2 years ago
Given right triangle DEF, what is the value of tan(F)?
Y_Kistochka [10]

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The answer is C.) 40/9

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3 years ago
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A publisher reports that 63c% of their readers own a personal computer. A marketing executive wants to test the claim that the p
rusak2 [61]

Answer:

Null hypothesis is not rejecting at 0.05 level.

Step-by-step explanation:

We are given that a publisher reports that 63% of their readers own a personal computer. A random sample of 110 found that 56% of the readers owned a personal computer.

And, a marketing executive wants to test the claim that the percentage is actually different from the reported percentage, i.e;

Null Hypothesis, H_0 : p = 0.63 {means that the percentage of readers who own a personal computer is same as reported 63%}

Alternate Hypothesis, H_1 : p \neq 0.63 {means that the percentage of readers who own a personal computer is different from the reported 63%}

The test statistics we will use here is;

                T.S. = \frac{\hat p -p}{\sqrt{\frac{\hat p(1- \hat p)}{n} } } ~ N(0,1)

where, p = actual % of readers who own a personal computer = 0.63

            \hat p = percentage of readers who own a personal computer in a

                   sample of 110 = 0.56

           n = sample size = 110

So, Test statistics = \frac{0.56 -0.63}{\sqrt{\frac{0.56(1- 0.56)}{110} } }

                             = -1.48

Now, at 0.05 significance level, the z table gives critical value of -1.96 to 1.96. Since our test statistics lie in the range of critical values which means it doesn't lie in the rejection region, so we have insufficient evidence to reject null hypothesis.

Therefore, null hypothesis is not rejecting at 0.05 level.

3 0
2 years ago
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