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2 years ago

Can someone help me with this?

Mathematics

1 answer:

KonstantinChe [14]2 years ago

4 0

Answer:

1. AE= 312

2. AC = 161

Step-by-step explanation:

Hey there!

In order to solve the first question, we need to think of possible "C" values

Then after finding the "C" value, we can then find the "A" and "E" values

So:

AC = 24 and CE = 13

What factors do 24 and 13 have in common?

Well, only 1 since 13 is a prime number]

So this means that C has to be 1

Now we know that A is 24 and E is 13

Now we have to multiply them together

24 x 13 is 312

So AE= 312

Now we go on to the second question

We can use the same strategy we used for number one

Instead of finding what C is, we need to find out what E is, since there is E in both of the statements

So:

CE = 7 and AE = 23

What common factors do these two numbers have?

Again, only 1, meaning that E can only be 1

Now we know that C is 7 and A is 23

Since we know this, we can multiply these two numbers together and get our final answer: 161

So AC = 161

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mina [271]

Answer:

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Step-by-step explanation:

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3 years ago

An engineer is comparing voltages for two types of batteries (K and Q) using a sample of 75 type K batteries and a sample of 46

lorasvet [3.4K]

The decision rule for rejecting the null hypothesis, considering the t-distribution, is of:

|t| < 1.9801 -> do not reject the null hypothesis.

|t| > 1.9801 -> reject the null hypothesis.

<h3>What are the hypothesis tested?</h3>

At the null hypothesis, it is tested if there is not enough evidence to conclude that the mean voltage for these two types of batteries is different, that is, the subtraction of the sample means is of zero, hence:

$H_0: \mu_1 - \mu_2 = 0$

At the alternative hypothesis, it is tested if there is enough evidence to conclude that the mean voltage for these two types of batteries is different, that is, the subtraction of the sample means different of zero, hence:

$H_1: \mu_1 - \mu_2 \neq 0$

We have a two-tailed test, as we are testing if the mean is different of a value.

Considering the significance level of 0.05, with 75 + 46 - 2 = 119 df, the critical value for the test is given as follows:

|t| = 1.9801.

Hence the decision rule is:

|t| < 1.9801 -> do not reject the null hypothesis.

|t| > 1.9801 -> reject the null hypothesis.

More can be learned about the t-distribution in the test of an hypothesis at brainly.com/question/13873630

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1 year ago

Can anyone one help me with this question

n200080 [17]

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3 years ago

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Step-by-step explanation:

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