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3 years ago

Which of the following shows why the commutative property doesn't work under subtraction

Mathematics

2 answers:

Annette [7]3 years ago

5 0

Hello! To answer your question, you first need to know what the commutative property is and does. Down to the basics, it lets you switch around numbers and it would allow me to say that 2+5=7 and 5+2=7. This property also works in multiplication, because if I take 4*2=8, and 2*4 still equals 8. Now that I've either refreshed or hopefully helped you to understand the property better, why wouldn't it work under subtraction? If you switch numbers around in subtraction, will your answer still be the same? Once you answer that, take a peak at the worksheet! I hope this has helped in some way, and if you have any further questions send me a question or another message. Good luck :)

siniylev [52]3 years ago

3 0

Answer with explanation:

Commutative property under subtraction:

If a and b are any two real numbers such that,⇒ a-b =b-a, then we say that ,a and b Satisfy Commutative property under Subtraction.

Take any two real numbers

a=5.6

And,b=7

→a-b

=5.6-7

= -1.4

→b-a

=7-5.6

=1.4

So,a-b≠b-a

Therefore , we can say that, real numbers doesn't satisfy commutative property under subtraction.

Option D: ⇒5-1≠1-5

This statement shows that, commutative property doesn't work under subtraction.

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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).

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