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3 years ago

What is 14.54 - n = 7.52?

2 answers:

6 0

The answer is 7.02 because you subtract 14.54 over to 7.52 and get a negative -7.02 but you can´t have a -n so you have to divide -7.02/-n and it goes postive so
n=7.02
Hope That Helps :)

navik [9.2K]3 years ago

6 0

If you are trying to figure what the value of n is, it would be 7.02 because you have to isolate the variable (n) by subtracting 14.54 on both sides giving you
-n=-7.02 and then you divide both sides with -1 because it is known that a variable by itself is multiplied by one, -1 in this case because the negative sign was placed in front of the variable.

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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:

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