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katen-ka-za
2 months ago
6

Solve each equation for the indicated variable abc = 1/2; b

Mathematics
2 answers:
Zina [12.3K]2 months ago
6 0
Kindly rephrase your question so it is clear. Thank you and best wishes! :D
Zina [12.3K]2 months ago
0 0

First, divide both sides by b to get ac = \(\frac{1}{2b}\). Then multiply both sides by 2, leading to
2ac = \(\frac{1}{b}\).

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The UF athletic department wanted to determine if students came to more than 4 home football games on average per season. They r
Svet_ta [12734]
The correct response is D; we can treat the survey as one involving a single variable, responding with either "yes, I attended more than 4 games" or "no." Given the random survey of 10 students meets the randomization criterion and that their responses are independent, options B and C can be dismissed. However, since only 10 students were surveyed, the confidence interval will not be narrow. As per Statistical Solutions, a minimum of 10 subjects per variable is essential for regression analysis. If the query concerns the number of games each student attended, the potential variables increase; conversely, if it solely asks, “Did you attend more than 4 games?”, then we only consider a single variable, making 10 students sufficient.
5 0
1 month ago
The Insure.com website reports that the mean annual premium for automobile insurance in the United States was $1,503 in March 20
PIT_PIT [12445]

Answer:

a) Null and alternative hypothesis

H_0: \mu=1503\\\\H_a:\mu< 1503

b) Point estimate d = -$78

c) Test statistic t = -2.438

P-value = 0.0113

Reject H0. This indicates that the average automobile premium in Pennsylvania is lower than in the nation.

Step-by-step breakdown:

This is a statistical test for the average population mean.

The hypothesis posits that car insurance in Pennsylvania is notably less expensive compared to the national average.

Accordingly, the null and alternative hypotheses are:

H_0: \mu=1503\\\\H_a:\mu< 1503

The significance level is set at 0.05.

The sample size is n=25.

The sample mean equates to M=1425.

A point estimate of the difference between the Pennsylvania mean premium and the national average can be computed using the sample mean:

d=M-\mu=1425-1503=-78

Given that the standard deviation of the population is unknown, we approximate it using the sample standard deviation, which is s=160.

The estimated standard error of the mean is determined with the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{160}{\sqrt{25}}=32

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{1425-1503}{32}=\dfrac{-78}{32}=-2.438

The degrees of freedom for this sample size stand at:

df=n-1=25-1=24

This constitutes a left-tailed test, with 24 degrees of freedom and t=-2.438, rendering the P-value as (per a t-table):

\text{P-value}=P(t

As the P-value (0.0113) falls below the significance level (0.05), the results prove significant.

Thus, the null hypothesis obtains dismissal.

At a 0.05 significance level, there's sufficient evidence to assert that car insurance in Pennsylvania costs notably less than the national average.

8 0
1 month ago
Jonathan has 3/4 pound of grapes.how many 1/8pound servings can Jonathan make from his grapes
Inessa [12570]

Response:

6 servings.

Detailed explanation:

To determine this, divide 3/4 by 1/8, here are the steps:

3/4 divided by 1/8

3/4 times 8/1

(3 × 8 = 24); (4 × 1 = 4)

24 divided by 4 results in 6.

I hope this assists! If possible, please grant me brainliest, as I need it to level up! Thank you.

5 0
1 month ago
The value of y at x=-1in the equation 5y=2 is
Zina [12379]

Answer: y = 2/5 --> y = 0.4

Step-by-step explanation:

The equation 5y = 2 represents a horizontal line at y = 2/5.

This means that the y-coordinate holds the value of 2/5 = 0.4 for any x-coordinate.

7 0
2 months ago
A 95% confidence interval for the population proportion of professional tennis players who earn more than 2 million dollars a ye
Svet_ta [12734]

Response:

e. 545

Detailed explanation:

In a survey sample containing n individuals, with a success probability of \pi, and a confidence level of 1-\alpha, the ensuing confidence interval for proportions is established.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

Wherein

z denotes the z-score corresponding to a probability value of 1 - \frac{\alpha}{2}.

For this scenario, we find:

The estimate averages the two bounds. Thus \pi = \frac{0.82+0.88}{2} = 0.85

95% confidence level

Consequently, z represents the z value corresponding to the p-value of 1 - \frac{0.05}{2} = 0.975, hence Z = 1.96.

The lower boundary of this interval is:

L = \pi - z\sqrt{\frac{\pi(1-\pi)}{n}}

In this query, L = 0.82. Therefore

0.82 = 0.85 - 1.96\sqrt{\frac{0.85*0.15}{n}}

1.96\sqrt{\frac{0.85*0.15}{n}} = 0.03

0.03\sqrt{n} = 1.96\sqrt{0.85*0.15}

\sqrt{n} = \frac{1.96\sqrt{0.85*0.15}}{0.03}

(\sqrt{n})^{2} = (\frac{1.96\sqrt{0.85*0.15}}{0.03})^{2}

n = 544.23

Thus, the accurate response is:

e. 545

4 0
13 days ago
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