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anzhelika [568]
3 years ago
15

Solve for x. 38(2x+16)−2=13

Mathematics
2 answers:
Lina20 [59]3 years ago
6 0

76x+608-2=13

76x+606=13

76x= -593

x= -7.8

Anastaziya [24]3 years ago
3 0

How to get answer:

Step 1: Simplify both sides of the equation.

38(2x+16)−2=13

(38)(2x)+(38)(16)+−2=13(Distribute)

76x+608+−2=13

(76x)+(608+−2)=13(Combine Like Terms)

76x+606=13

76x+606=13

Step 2: Subtract 606 from both sides.

76x+606−606=13−606

76x=−593

Step 3: Divide both sides by 76.

\frac{76x}{76} = \frac{-593}{76}

Answer:

Decimal Form: −7.80263157…

Fraction Form: \frac{-593}{76}

Mixed Number Form: -7\frac{61}{76}

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A random sample of 40 binomial trials resulted in 14 successes. test the claim that the population proportion of successes does
mylen [45]

Using the z-distribution, since the p-value of the test is of 0.057 > 0.05, there is not enough evidence that the population proportion of successes does not equal 0.50.

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

At the null hypotheses, we test if the proportion of successes equals 0.5, hence:

H_0: p = 0.5

At the alternative hypotheses, we test if it does not equal, hence:

H_1: p \neq 0.5

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

The test statistic is given by:

z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}

In which:

  • \overline{p} is the sample proportion.
  • p is the proportion tested at the null hypothesis.
  • n is the sample size.

For this problem, the parameters are given by:

n = 40, \overline{p} = \frac{14}{40} = 0.35, p = 0.5

Hence the test statistic is given by:

z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}

z = (0.35 - 0.5)/(0.5/sqrt(40))

z = -1.9.

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

Using a z-distribution calculator, considering a two-tailed test, as we are testing if the proportion is different of a value, with z = -1.9, we get that the p-value of the test is of 0.057.

Since the p-value of the test is of 0.057 > 0.05, there is not enough evidence that the population proportion of successes does not equal 0.50.

More can be learned about the z-distribution at brainly.com/question/16313918

#SPJ4

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