A farmer estimates that he has 9,000 bees producing honey on his farm. The farmer becomes concerned when he realizes the populat

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:

The equation (in t(n) notation) can be expressed as t(n) = 325(1.04) raised to the 12th power. Alternatively, you would obtain 520.34 if rounding up the half cent.

Step-by-step explanation:

The initial value is 325. The factor of 1.04 acts as the multiplier since the growth rate is 0.04, and including 1 is necessary to prevent a 96% decline each iteration.

Sample Response: Rotations fall under rigid transformations, meaning they keep the figure’s size, lengths, shapes, and angle measurements unchanged. Despite this, the orientation of the figure does change. Line segments that connect the center of rotation to points on both the original figure and its rotated image are congruent in length. However, corresponding edges between matching vertices do not run parallel to each other.

Response:

Step-by-step clarification:

Reply:

a) y-8 = (y₀-8), b) 2y -5 = (2y₀-5)

Clarification:

To address these equations, using direct integration is the simplest approach.

a) The equation provided is

          dy / dt = -y + 8

         dy / (y-8) = dt

We substitute variables

          y-8 = u

         dy = du

Substituting and integrating gives us

           ∫ du / u = ∫ dt

           Ln (y-8) = t

Evaluating at the lower limits t = 0 for y = y₀

          ln (y-8) - ln (y₀-8) = t-0

Simplifying the equation results in

           ln (y-8 / y₀-8) = t

           y-8 / y₀-8 =

            y-8 = (y₀-8)

b) The equation here is

            dy / dt = 2y -5

            u = 2y -5

            du = 2 dy

            du / 2u = dt

Integrating now

             ½ Ln (2y-5) = t

Evaluating at limits gives

            ½ [ln (2y-5) - ln (2y₀-5)] = t

            Ln (2y-5 / 2y₀-5) = 2t

            2y -5 = (2y₀-5)

c) The equation here bears a strong resemblance to the previous one

             u = 2y -10

             du = 2 dy

             ∫ du / 2u = dt

             ln (2y-10) = 2t

We evaluate this to get

             ln (2y-10) –ln (2y₀-10) = 2t

               2y-10 = (2y₀-10)