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3 years ago

A rectangle is 4 units long and 2 units wide. What is its area?

Mathematics

2 answers:

Irina18 [472]3 years ago

7 0

rectangle area = l * w

2 * 4 =

8 units²

nordsb [41]3 years ago

5 0

8 square units formula (length x with = area) tip in area the units are square 4x2=8 square units

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Pants $80 buy one at regular pric. Get a second one for half price. (How much for 2 pair?)

castortr0y [4]

Answer:

$200 will be price of 2 pairs

Step-by-step explanation:

It is given that the price of one pant is $80

The second is given at half price which is $40

but in the case of pants buying or anything buying when we are given an off then after buying an item the same discount is given for all

so as he bought one pant then the other pant will be given at half price

now the third pant will also be given at half price i.e: $40

fourth price will also be given at $40

so four pants will become 2 pairs

price of 2 pairs = $80+$40+$40+$40

=$200

5 0

3 years ago

Consider the following data set. The data were actually collected in pairs, and each row represents a pair.

Anna11 [10]

Answer:

Step-by-step explanation:

Hello!

Using the data sets you have to analyze them as if they are two independent samples.

The hypotheses are:

H₀: μ₁=μ₂

H₁: μ₁≠μ₂

α: 0.05

Assuming that both data sets are from a normal distribution and both population variances, although unknown, are equal, the statistic to use is:

$t=\frac{(X[bar]_1-X[bar]_2)-(Mu_1-Mu_2)}{Sa\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } ~~t_{n_1+n_2-2}$

$Sa^2= \frac{(n_1-1)S^2_1+(n_2-1)S^2_2}{n_1+n_2-2}$

$Sa^2= \frac{9*5.38+9*3.65}{10+10-2} =4.51$

Sa= 2.12

$t_{H_0}= \frac{(49.69-50.52)-0}{2.12*\sqrt{\frac{1}{10} +\frac{1}{10} } } = -0.875$

The p-value for the two-tailed test is:

P( $t_{18}$ ≤-0.875) + P( $t_{18}$ ≥0.875)= P( $t_{18}$ ≤-0.875) + (1 - P( $t_{18}$ ≤0.875))= 0.1965 + ( 1 - 0.8035)= 0.393

Since there is no significant level I determined it at 5%, comparing it to the p-value, the hypothesis test is not significant. Meaning that the difference between the population means of both groups is equal to zero.

Considering the given data as a paired sample, you have to determine the variable difference to conduct the test:

Xd: the difference between X₁ and X₂

Before the sample mean and sample standard deviation you have to calculate the difference between the observations of group 1 and group 2. (2nd attachment)

Mean X[bar]d= -0.83

Standard deviation Sd= 1.28

The hypotheses for the paired sample test are:

H₀: μd=0

H₁: μd≠0

α: 0.05

$t= \frac{X[bar]d-Mud}{\frac{Sd}{\sqrt{n} } } ~~t_{n-1}$

$t_{H_0}= \frac{-0.83-0}{\frac{1.28}{ \sqrt{10} }} = -2.05$

The p-value for the two-tailed test is:

P( $t_{9}$ ≤-2.05) + P( $t_{9}$ ≥2.05)= P( $t_{9}$ ≤-2.05) + (1 - P( $t_{9}$ ≤2.05))= 0.0353 + (1 - 0.9647)= 0.0706

Comparing the p-value with the significance level, the hypothesis test is not significant. Meaning that the population mean of the difference between group 1 and group 2 is equal to cero. There is no difference between the two groups.

The value of the statistic in "a" is greater than the value of the statistic obtained in "b". Since there are two samples used an "a" the degrees of freedom of the test are the double as the ones used in "b". The p-value obtained in "b" is around half of the p-value of "a".

The main difference is that in "a" you compared two different samples but in "b" you compared two dependent samples when analyzing paired data the "effect of the individual" is removed from the equation.

I hope this helps!