What is the value of the expression? 84-24428 13 115 13.4 152 O

6,000,000 How do say this number

Nataly [62]

Answer:

six million

Step-by-step explanation:

4 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

3 pumps, working 8 hours a day, can empty a tank in 2 days. how many hours a day must 4 pumps work to empty the tank in 1 day?

nexus9112 [7]

It takes 3 pumps 16 hours to empty a tank
Meaning it takes 48 hours for one pump by its self
So if there are 4 pumps
48/4 it would take them 12 hours
They would need to work 12 hours to do it in 1 day

4 0

3 years ago

Select the correct answer. Thank you

Setler79 [48]

Answer:

I think a

Step-by-step explanation:

7 0

3 years ago

Can anyone answer these?

soldier1979 [14.2K]

Answer:

4) 3 for the numerator. 9 for the denominator

5) 6 for the first blank square and 48 for the second blank.

6) 2 for the denominator. 6 for the numerator. 30 for the second numerator.

Step-by-step explanation:

sorry but those are the only ones i know

Hope this helps!

3 0

2 years ago