Enter a recursive rule for the geometric sequence. 6, −18, 54, −162, ...

Answer:

At the α = 0.10 level, there is no substantial evidence indicating that the average vertical jump for students at this school differs from 15 inches.

Step-by-step explanation:

A hypothesis test is necessary to verify the assertion that the average vertical jump of students diverges from 15 inches.

The null and alternative hypotheses are:

The significance level is set at 0.10.

The sample mean recorded is 17, and the sample standard deviation is 5.37.

The degrees of freedom are calculated as df=(20-1)=19.

The t-statistic is:

The two-tailed P-value corresponding to t=1.67 is P=0.11132.

<pSince this P-value exceeds the significance level, the result is not significant. Therefore, the null hypothesis remains unchallenged.

At the α = 0.10 level, there is no compelling evidence that the average vertical jump of students at this school deviates from 15 inches.

The first equation is x + y = 29, and the second is 5x + 2y = 100.

x=88. The third line serves as a transversal, meaning that 92 and x are supplementary angles.

Answer:

We conclude that less than or equal to 50% of the student body supports him significantly.

Step-by-step explanation:

Mike, while campaigning for the student government presidency, is keen to determine if more than 50% of the student body supports him.

A random sample of 100 students was surveyed, with 55 expressing support for Mike.

Let p = the proportion of students backing Mike.

Thus, Null Hypothesis, : p 50% {indicating that the proportion of supporters among the student body is significantly less than or equal to 50%}

Alternate Hypothesis, : p > 50% {indicating that the proportion of supporters among the student body is significantly more than 50%}

The test statistics to be utilized here One-sample z proportion statistics;

T.S. = ~ N(0,1)

where, = the sampled proportion in favor of Mike = = 0.55

n = number of sampled students = 100

Therefore, test statistics =

= 1.01

The z test statistic is 1.01.

At a significance level of 0.05, the z table provides a critical value of 1.645 for a right-tailed test.

Given that our test statistic falls below the critical value of z (1.01 < 1.645), we lack sufficient evidence to reject the null hypothesis as it remains outside the rejection region, leading to failure to reject our null hypothesis.

As a result, we conclude that less than or equal to 50% of the student body is in favor of him or the proportion supporting Mike is not significantly greater than 50%.

Given:
1 pack = 5 pencils and cardboard.
1 pack should weigh between 60 grams and 95 grams

60g < x < 95g; where x signifies 1 pack.

Cardboard: 15 grams.

95g - 15g = 80g represents the maximum total weight of 5 pencils.
80g / 5 = 16g is the maximum weight for a single pencil.

60g - 15g = 45g is the minimum total weight of 5 pencils.
45g / 5 = 9g is the minimum weight for a single pencil.

9 < x < 16; where x represents a single pencil in the pack.