The organic garden at a restaurant is rectangular in shape and has a perimeter of 84ft. If the length of the garden is 24ft long

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%.

Answer:

Choice A.

Detailed explanation:

The provided expression is

where, .

Our goal is to identify an expression that matches the one given.

The current expression can be expressed as

Thus, the right choice is A.

Answer: Approximately 60%

Step-by-step explanation:

Answer:

To measure 1.25 mL of dye solution, use 5 mL of dye combined with 3 mL of water

Consequently, the total solution volume reaches 8 mL.

Next,

This 8 mL solution consists of

part water.In the following step, extract 2 mL of this solution.

Thus,

The dye content in the 2 mL solution is

or

The 2 mL solution contains 1.25 mL of dye

Hence,

the required 1.25 mL of dye is now accurately measured.

Given:

A 10-mL graduated cylinder calibrated in 1-mL increments

The dye solution to be measured = 1.25 mL

Now,

Combine 5 mL of dye with 3 mL of water

This creates a total solution volume of 8 mL.

Next,

This 8 mL solution comprises

part water.Then, extract 2 mL from this solution.

Thus,

The concentration of dye in the 2 mL solution measures

or

1.25 mL of dye per 2 mL solution