When Joanne checked her Work by substituting the value of the variable back into the equation she got an answer of 12.34 equals

The mistake lies in the fact that the logarithms have different bases. The one-to-one property of logarithms cannot be applied unless the bases are identical. <span>To correct this, the change of base formula should be used to express the logarithms with a uniform base.
I have confirmed this using Edge.</span>

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%.

A. Mean and standard deviation.

The sampling distribution’s mean closely matches the population mean. Since the population mean is 174.5, the sampling distribution’s mean equals this value.

The standard deviation of the sampling distribution is:

            σₓ̄ = σ / √n

Plugging in values:

            σₓ̄ = 6.9 / √25 = 1.38

b. Calculate z-scores for both values:
         
      z = (value - mean) / standard deviation

For 172.5:
     z = (172.5 - 174.5) / 1.38 = -1.49
Corresponding probability ≈ 0.068

For 175.8:
   z = (175.8 - 174.5) / 1.38 = 0.94
Corresponding probability ≈ 0.83

The difference between these probabilities is 0.762.

Approximately 0.762 × 200 = 152 sample means lie between 172.5 and 175.8.

c. Z-score for 172 cm:

    z = (172 - 174.5) / 1.38 = -1.81
Probability ≈ 0.03

Hence, samples with means below 172 cm equal 0.03 × 200 = 6.

a) The area of a rectangle is calculated by multiplying the length by the width:
   A = l·w

This formula allows us to express the width in terms of area and length. By dividing A by l, we find
   A/l = w

We also recognize that the perimeter of a rectangle is the total length around it.
   P = l + w + l + w
   P = 2(l + w)
We aim to rewrite the perimeter formula to isolate l on one side, using the expression for w derived earlier.
   P = 2(l + A/l)

By substituting the known value for A, we can express p(l) as
   p(l) = 2(l + 25/l)
   p(l) = 2l + 50/l


b) For lengths exceeding widths, we have
   l > w
   l > A/l
   l² > A
   l > √A
   l > √25

This indicates that the domain of p(l) is
   l ∈ (5, ∞)..... meters

Answer:

Step-by-step breakdown:

The necessary formula for this problem is

which resolves to

leading to

36 + 6x = 40 + 5x, and consequently

x = 4

Thus, DG equals 5 + 4 + 3, resulting in 12