Help me out guys for math class

I’ll mark you brainlist I’ll mark you brainlist

Step2247 [10]

Answer:

Step-by-step explanation:

Slope of the line =3\2

7 0

3 years ago

Read 2 more answers

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%.

(c) The necessary assumption is that the population is normally distributed.

(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

Thank you so much for helping meeeeee

Ivenika [448]

Hello, and thanks for posting your question!

A function would be a set of ordered pairs that have no repeating "x" values, look at the example below. (V)

{(1, 0), (2, 0), (3, 0)} = Function
{(1, 0), (1, 0), (2, 0)} = Not a function.

According to this information, your answer is C. Both Set A and Set B.

Hope this helps! ☺♥

7 0

3 years ago

I'm so confused please helpppppppp How is C correct please show your work

777dan777 [17]

Answer:

Step-by-step explanation:

She makes $325 a week, so go to the 8th row, for weekly income of $300 to $400. She claims 2 exemptions, so go to the third column labeled 2. The value is 2. That means the $2 is withheld from her paycheck on top of the 6.25% for FICA. The total is therefore:

$2 + (0.0625) ($325)

$22.3125

Rounded to the nearest cent, that's $22.31.

3 0

3 years ago

NEED HELP QUICK PLEASE !!

xxMikexx [17]

Answer:

you have already choose correct one's

BC=DC

Step-by-step explanation:

6 0

3 years ago