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3 years ago

To solve 6x=72 do you use the addition property of equality

Mathematics

1 answer:

lawyer [7]3 years ago

3 0

Answer:

No , the division property of equality.

Step-by-step explanation:

The Division Property of Equality states that if you divide both sides of an equation by the same nonzero number, the sides remain equal.

Hope this helps!!

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Jason's two test scores in English are 87 and 91. What must he score on the third test to get an average test score of 92 or bet

irga5000 [103]

You first find the average of the first two test scores. When you find the average, you add the two numbers together and then divide by two which gives you 89. You then plug in different scores above the number 89 and find the average of those (89 and x) to see which gives you an average of 92. It’s mostly a guess and check type of problem.

5 0

3 years ago

Please help me with this please actually help me please and thank you

Westkost [7]

Answer:

C. (x, y) → (x, -y)

Step-by-step explanation:

The algebraic representation that correctly describes a reflection over the x-axis is (x, y) → (x, -y)

Here's an example to show understanding

Plot A is (4,6) and the reflection over the x-axis would be (4,-6)

8 0

1 year 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!