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Helen [10]
2 years ago
14

Is this statement true or false? a^(-3)+a^(-3)=a^(-6)​

Mathematics
2 answers:
EleoNora [17]2 years ago
7 0

It's false

..............

..............

..............

prohojiy [21]2 years ago
6 0
<h3>Answer: False</h3>

Reason:

It should be a^(-3)*a^(-3)=a^(-6)​ based on the rule a^b*a^c = a^(b+c)

A way to see why the equation a^(-3)+a^(-3)=a^(-6)​ is false is to pick a number like a = 1 to find that...

a^(-3)+a^(-3)=a^(-6)​

1^(-3)+1^(-3)=1^(-6)​

1/(1^3) + 1/(1^3) = 1/(1^6)

1/1 + 1/1 = 1/1

1+1 = 1

2 = 1

Which is false, so this shows the original equation is false for a = 1. Therefore it is false in general.

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If you really want to figure this out you make an equation to solve for the time
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7 0
3 years ago
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Simplify (-2 + 3i) + 2(5 + 6i). Enter your answer in the form a + bi.
Reptile [31]

We will have the following:

(-2+3i)+2(5+6i)=-2+3i+10+12i

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3 0
1 year ago
10. A manufacturer wanted to know if more coupons would be redeemed if they were mailed to the
Mandarinka [93]

Using the t-distribution, it is found that the data does not provide convincing evidence that the mean number is greater when the coupons were addressed to a female.

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

At the null hypothesis, it is tested if there is no difference, that is, the mean is of 0, hence:

H_0: \mu = 0

At the alternative hypothesis, it is tested if the mean number is greater for females, that is, the mean is greater than 0, hence:

H_1: \mu > 0

<h3>What is the test statistic?</h3>

The test statistic is given by:

t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}

The parameters are:

  • \overline{x} is the sample mean.
  • \mu is the value tested at the null hypothesis.
  • s is the standard deviation of the sample.
  • n is the sample size.

In this problem, the parameters are given as follows:

\overline{x} = 1.5, \mu = 0, s = 4.75, n = 50.

Hence, the test statistic is given by:

t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}

t = \frac{1.5 - 0}{\frac{4.75}{\sqrt{50}}}

t = 2.23

<h3>What is the conclusion?</h3>

Considering a <em>right-tailed test</em>, as we are testing if the mean is greater than a value, with a <em>significance level of 0.01 and 50 - 1 = 49 df</em>, the critical value is given by t^{\ast} = 2.4.

Since the test statistic is less than the critical value for the right-tailed test, the data does not provide convincing evidence that the mean number is greater when the coupons were addressed to a female.

More can be learned about the t-distribution at brainly.com/question/26454209

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Answer:

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