Please help thank you so much if you do

Struck-by injuries are produced by forcible contact or impact between the injured person and ________________.

wariber [46]

C. an object or a piece of equipment :-)

6 0

2 years ago

Confidence Interval Mistakes and Misunderstandingsâ€”Suppose that 500 randomly selected recent graduates of a university were as

kvv77 [185]

Answer:

The correct 95% confidence interval is (8.4, 8.8).

Step-by-step explanation:

The information provided is:

$n=500\\\bar x=8.6\\\sigma=2.2$

(a)

The (1 - α)% confidence interval for population mean (μ) is:

$CI=\bar x\pm z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}$

The 95% confidence interval for the average satisfaction score is computed as:

8.6 ± 1.96 (2.2)

This confidence interval is incorrect.

Because the critical value is multiplied directly by the standard deviation.

The correct interval is:

$8.6\pm 1.96 (\frac{2.2}{\sqrt{500}})=8.6\pm 0.20=(8.4,\ 8.6)$

(b)

The (1 - α)% confidence interval for the parameter implies that there is (1 - α)% confidence or certainty that the true parameter value is contained in the interval.

The 95% confidence interval for the mean rating, (8.4, 8.8) implies that the true there is a 95% confidence that the true parameter value is contained in this interval.

The mistake is that the student concluded that the sample mean is contained in between the interval. This is incorrect because the population is predicted to be contained in the interval.

(c)

The (1 - α)% confidence interval for population parameter implies that there is a (1 - α) probability that the true value of the parameter is included in the interval.

The 95% confidence interval for the mean rating, (8.4, 8.8) implies that the true mean satisfaction score is contained between 8.4 and 8.8 with probability 0.95 or 95%.

Thus, the students is not making any misinterpretation.

(d)

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

In this case the sample size is,

n = 500 > 30

Thus, a Normal distribution can be applied to approximate the distribution of the alumni ratings.

7 0

3 years ago

How do i figure out [(3x²y³)2] ³

dybincka [34]

Answer:

216x⁶y⁹

Step-by-step explanation:

[(3x²y³)2]³

[6x²y³]³

[6]³.[x²]³.[y³]³

216.x⁶.y⁹

216x⁶y⁹

6 0

1 year ago

How do I do this please help

Umnica [9.8K]

Answer:

wrong as his calculation was incorrect at initial level itself when he found the result of the division

Step-by-step explanation:

Andre said

3 ÷ $\frac{2}{3}$

$=3 \times \frac{3}{2}$

$=\frac{3 \times 3}{2}$

$=\frac{9}{2}$

but andre calculated it as

$=4\tfrac{1}{3}$

$=\frac{4 \times 3 + 1}{3}$

$=\frac{13}{3}$

Hence his calculation was incorrect at initial level itself

4 0

3 years ago

A fitness club charges $25 per month as well as a one-time $80 fee of all new members which equation represents the cost in doll

madam [21]

Answer: So if you want to find the total cost (c) for a certain number of months (m) and it costs $35 per month, you just need to multiply the number of months by the cost per month. So

35m=c

If you want to see the units cancel

($35/month)(months)=$

The months will cancel out, and you will be left with total dollars.

Not sure about the second part of your question, but if you wanted to figure out how much it would cost to be a member for 6 months, you would just plug in 6 for m.

35(6)=c

c=210

So $210

Step-by-step explanation:

4 0

3 years ago