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Readme [11.4K]
3 years ago
15

Use the distributive property to simplify the expression 5(3 +x) – 2

Mathematics
2 answers:
Anna007 [38]3 years ago
7 0

Hello!

<em><u>Answer: 5x+13</u></em>

Explanation:

Distributive property: → a(b+c)=ab+ac

Expand the form.

5(3+x)

multiply by 5*3.

5*3=15

=15+5x

15+5x-2

Simplify by equation or expand.

15+5x-2=5x+13

Hope this helps!

insens350 [35]3 years ago
4 0

5(3 +x) – 2 = (5+3)*(5+x)

                       (8-2)*(5x-2)

                         6*5x-2

this Is what I came up with.

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Suppose in Loma, Montana the temperature rose from -2.3 degrees Fahrenheit to 8.6 degrees Fahrenheit. What is the change in temp
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A: 10.9

Step-by-step explanation:   -2.3 is 2.3 away from zero then add the length from zero to the postive number which in this case is 8.6, therefore the equation is 2.3 + 8.6 = 10.9

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A 30-foot pole is stabilized by a wire attached to the ground 10 feet from the base of the pole. how long is the wire?
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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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