What is (-1/3)+1/4+(-1/5)

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).

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Find the 22nd term. 3, 7.5, 12, 16.5

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Answer:

either or super sore

Step-by-step explanation:

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WORTH 50 POINTS Simplify -5a-6ab+10ab+2a please give explanation.

Ymorist [56]

Answer:

-3a + 4ab

Step-by-step explanation:

-5a + 2a = -3a

-6ab + 10ab = 4ab

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Mjhs is refinishing a basketball court and decided to paint everything shown below blue. How many square feet will be painted bl

SashulF [63]

Answer:

<h2>Mjhs needs 189.25 square feet of blue paint, approximately.</h2>

Step-by-step explanation:

As you can observe in the image attached, the basketball court is a composite figure formed by a rectangle and half of a circle.

The rectangle has dimensions of 10 feet by 15 feet. The circle has a diameter of 10 feet.

The expression of this composite area would be

$A= w \times l + \frac{\pi (\frac{l}{2} )^{2} }{2}$

Where $w = 15 \ ft$ and $l = 10 \ ft$ . Replacing these values, we have

$A= 15 \times 10 + \frac{3.14 (\frac{10}{2} )^{2} }{2}\\ A=150 + \frac{3.14 \times 25}{2}=150+39.25 \\ A \approx 189.25 \ ft^{2}$

Therefore, Mjhs needs 189.25 square feet of blue paint, approximately.

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Y+1=3(x-4)y+1=3(x−4) find the slope

julia-pushkina [17]

Answer:

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y+1= (3x-12)y+1=3x-12

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