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evablogger [386]
3 years ago
6

3 x (3^4) = 3x what is x

Mathematics
1 answer:
Llana [10]3 years ago
6 0

Answer:

Step-by-step explanation:

3 x 81 (to the power of 4 is the same as the square, squared, so 9 x 9) = 3x

243 = 3x

81 = x

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Answer:

\lim_{x \to 3} g(x) = DNE

Step-by-step explanation:

Observations:

  • As x approaches 3 from the left side, f(x) approaches -1.6
  • As x approaches 3 from the right side, f(x) approaches 1.6

Conclusion:

  • Because the right and left-side limits are not equal to each other, the limit does not exist
  • Therefore, \lim_{x \to 3} g(x) = DNE
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In simplifying the expression 2( 3x + 5) -6, what is the first thing to do?
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She would have to get 46 questions correct.
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The average undergraduate cost for tuition, fees, and room and board for two-year institutions last year was $13,252. The follow
kipiarov [429]

Answer:

The p-value of the test is 0.0041 < 0.05, which means that there is sufficient evidence at the 0.05 significance level to conclude that the mean cost has increased.

Step-by-step explanation:

The average undergraduate cost for tuition, fees, and room and board for two-year institutions last year was $13,252. Test if the mean cost has increased.

At the null hypothesis, we test if the mean cost is still the same, that is:

H_0: \mu = 13252

At the alternative hypothesis, we test if the mean cost has increased, that is:

H_1: \mu > 13252

The test statistic is:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question

t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, s is the standard deviation and n is the size of the sample.

13252 is tested at the null hypothesis:

This means that \mu = 13252

The following year, a random sample of 20 two-year institutions had a mean of $15,560 and a standard deviation of $3500.

This means that n = 20, X = 15560, s = 3500

Value of the test statistic:

t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}

t = \frac{15560 - 13252}{\frac{3500}{\sqrt{20}}}

t = 2.95

P-value of the test and decision:

The p-value of the test is found using a t-score calculator, with a right-tailed test, with 20-1 = 19 degrees of freedom and t = 2.95. Thus, the p-value of the test is 0.0041.

The p-value of the test is 0.0041 < 0.05, which means that there is sufficient evidence at the 0.05 significance level to conclude that the mean cost has increased.

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3 years ago
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