All categories

3 years ago

A club with 76 members is planning a camping trip They need to rent cabins and tents for sleeping.

1 answer:

4 0

The answer is B.4 because you need to divide 76 by 8 which will give you 9.5. Now, you can't have half of a person; so just cut it off. So divide that again by 2 and your answer is B.4 tents.

You might be interested in

The temperature of a certain solution is estimated by taking a large number of independent measurements and averaging them. The

Mademuasel [1]

Answer:

(a) The 95% confidence interval for the temperature is (36.80°C, 37.20°C).

(b) The confidence level is 68%.

(d) The 95% confidence interval for the temperature if 10 measurements were made is (36.93°C, 37.07°C).

Step-by-step explanation:

Let X = temperature of a certain solution.

The estimated mean temperature is, $\bar x=37^{o}C$ .

The estimated standard deviation is, $s=0.1^{o}C$ .

(a)

The general form of a (1 - α)% confidence interval is:

$CI=SS\pm CV\times SD$

Here,

SS = sample statistic

CV = critical value

SD = standard deviation

It is provided that a large number of independent measurements are taken to estimate the mean and standard deviation.

Since the sample size is large use a z-confidence interval.

The critical value of z for 95% confidence interval is:

$z_{0.025}=1.96$

Compute the confidence interval as follows:

$CI=SS\pm CV\times SD\\=37\pm 1.96\times 0.1\\=37\pm0.196\\=(36.804, 37.196)\\\approx (36.80^{o}C, 37.20^{o}C)$

Thus, the 95% confidence interval for the temperature is (36.80°C, 37.20°C).

(b)

The confidence interval is, 37 ± 0.1°C.

Comparing the confidence interval with the general form:

$37\pm 0.1=SS\pm CV\times SD$

The critical value is,

CV = 1

Compute the value of P (-1 < Z < 1) as follows:

$P(-1$

The percentage of z-distribution between -1 and 1 is, 68%.

Thus, the confidence level is 68%.

(c)

The confidence interval for population mean can be constructed using either the z-interval or t-interval.

If the sample selected is small and the standard deviation is estimated from the sample, then a t-interval will be used to construct the confidence interval.

But this will be possible only if we assume that the population from which the sample is selected is Normally distributed.

Thus, the necessary assumption is that the population is normally distributed.

(d)

For n = 10 compute a 95% confidence interval for the temperature as follows:

The (1 - α)% t-confidence interval is:

$CI=\bar x\pm t_{\alpha/2, (n-1)}\times \frac{s}{\sqrt{n}}$

The critical value of t is:

$t_{\alpha/2, (n-1)}=t_{0.025, 9}=2.262$

*Use a t-table for the critical value.

The 95% confidence interval is:

$CI=37\pm 2.262\times \frac{0.1}{\sqrt{10}}\\=37\pm 0.072\\=(36.928, 37.072)\\\approx (36.93^{o}C, 37.07^{o}C)$

Thus, the 95% confidence interval for the temperature if 10 measurements were made is (36.93°C, 37.07°C).

3 0

3 years ago

In the equation 3x - 9 = 3, what is the first step we should do?

zavuch27 [327]

Anse

draw the line down the equal sign

Step-by-step explanation:

4 0

3 years ago

2. What is the tangent ratio of angle & within a right triangle?

Dmitry [639]

$▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪$

According to Trigonometry,

$\tan( \theta) = \dfrac{opposite }{adjacent}$

3 0

3 years ago

Read 2 more answers

Compute Δy and dy for the given values of x and dx = Δx. (Round your answers to three decimal places.) Compute Δy and dy for the

Darina [25.2K]

Base on the data you have given and with my calculation, i think the best value for Delta X and Dy is delta x = 2 and the DY = 1/2.I hope you are satisfied with my answer and feel free to ask for more if you have more clarifications and further questions

3 0

3 years ago

Read 2 more answers

Solve for j - algebra

hodyreva [135]

Answer:

k=57

Step-by-step explanation:

Multiply both sides by 5

k-2=55

Move the constant to the right-hand side and change its sign

k=55+2

k=57

8 0

3 years ago