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3 days ago

Hugger Polls contends that an agent conducts a mean of 53 in-depth home surveys every week. A streamlined survey form has been i

ntroduced, and Hugger wants to evaluate its effectiveness. The number of in-depth surveys conducted during a week by a random sample of 15 agents are: 53 57 50 55 58 54 60 52 59 62 60 60 51 59 56 Picture Click here for the Excel Data File At the .05 level of significance, can we conclude that the mean number of interviews conducted by the agents is more than 53 per week? (Round your answer to 3 decimal places.) Reject H0 : μ ≤ 53 when the test statistic is . Compute the value of the test statistic. (Round your answer to 3 decimal places.) Value of the test statistic What is your decision regarding H0? Reject Do not reject Estimate the p-value. The p-value is .

Mathematics

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for the level 3 course, examination hours cost twice as much as workshop hours and workshop hours cost twice as much as lecture

Leona [12618]

Answer:

The hourly rate for lectures is $7.33

Step-by-step explanation:

* Let's break down how to tackle the problem.

- For the level 3 course, examination hours are priced at double that of workshop hours.

- Workshop hours cost twice the rate of lecture hours.

- The total includes examination, workshop, and lecture hours.

- Examination lasts 3 hours, workshops 24 hours, and lectures 12 hours.

* Let’s denote the cost of lecture hours as $x per hour.

∴ The lectures cost $x per hour.

∵ Workshop charge is twice that of lectures

∴ Workshop hours cost 2(x) = 2x per hour.

∵ Examination fees are double that of workshop hours

∵ The workshop cost is 2x

∴ Examination fees are 2(2x) = 4x per hour.

- Combining costs for level 3 gives us the total of lecture, workshop, and examination hours.

∵ 12 hours for lectures

∵ 24 hours for workshops

∵ 3 hours for examinations

∵ Thus the total cost for level 3 = 12(x) + 24(2x) + 3(4x).

∴ Total cost for level 3 = 12x + 48x + 12x.

∵ Therefore, total cost = $528.

∴ 12x + 48x + 12x = 528.

∴ 72x = 528; hence we divide both sides by 72.

∴ x = 7.33.

∵ x represents the cost of lecture hours per hour.

∴ Therefore, the hourly price for lectures is $7.33.

6 0

3 months ago

Construct an interval estimate for the given parameter using the given sample statistic and margin of error. For μ1-μ2, using x¯

zzz [12365]

Answer: CI = (0, 8)

Step-by-step explanation: The confidence interval for the difference in means is given as

Lower limit

= (x1 - x2) + margin of error

Upper limit

= (x1 - x2) - margin of error

Where x1 - x2 = 8 and the margin of error = 8

For the lower limit,

= 8 - 8 = 0

For the upper limit

= 8 + 8 = 16

Thus, CI = (0, 8)

4 0

2 months ago

: Xem xét 2 nhà đầu tư A và B với đường cầu về cổ phiếu như sau: Nhà đầu tư A: 2P = 300 – 2Q Nhà đầu tư B: P = 150 – 3Q a. Tại m

Zina [12379]

Đáp án:

Giải thích từng bước:

5 0

2 months ago

Erika is writing a coordinate proof to show that the diagonals of a rectangle are congruent. She begins by assigning coordinates

Leona [12618]

Answer:

Option C is the right choice.

Step-by-step explanation:

The given coordinates define a rectangle, and our objective is to show that the diagonals JL and KM are congruent.

We know that rectangles possess four right angles.

To prove the congruence of the diagonals JL and KM, we will utilize the Pythagorean theorem.

In triangle KLM, KL has a length of b units while LM has a length of a units. By applying the Pythagorean theorem $\sqrt(a^{2}+b^{2})=KM$

In triangle JML, JM is b units long, and LM remains a units long. We again can apply the Pythagorean theorem $\sqrt(a^{2}+b^{2})=JL$

Thus, we find that $\sqrt(a^{2}+b^{2})=KM=LM$ and option C is the correct choice.

7 0

2 months ago

Read 2 more answers

A 16\dfrac1216 2 1 16, start fraction, 1, divided by, 2, end fraction kilometer stretch of road needs repairs. Workers can rep

lawyer [12517]

Answer:

The repair of the stretch of road will require

weeks to finish. $7\frac{1}{3}$

Step-by-step explanation:

Provided:

Length of the roads needing repair =

Length of the road workers can fix in a week = $16\frac{1}{2}\ km$

To ascertain how many weeks it will take to mend this road stretch. $2\frac{1}{4}\ km$

Solution:

Worker's road repair rate = per week.

Overall road length to repair =

$2\frac{1}{4}\ km$

The time required to repair $16\frac{1}{2}\ km$

of road by the workers can be calculated: $16\frac{1}{2}\ km$ ⇒

When dividing mixed numbers, we change them first into fractions. $\frac{16\frac{1}{2}\ km}{2\frac{1}{4}\ km}$

To compute fraction division, we take the reciprocal of the divisor and substitute division with multiplication:

⇒ $\frac{\frac{33}{2}} {\frac{9}{4}}$

We revert the fraction back into a mixed number.

Thus, it will take

weeks to complete the road repair task. $\frac{33}{2}\times \frac{4}{9}$

7 0

2 months ago