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4 years ago

5.5 = -2 + d

2 answers:

7 0

The value of <em>d</em> is 7.5.

The key to solving equations is doing the inverse, or opposite, function/process.
For example, you subtract when d + 1 = 2:
d + 1 = 2
(d + 1) -1 = (2) - 1
d = 1

In the same way, you add values when you have a negative:
5.5 = -2 + d = d - 2
5.5 = d - 2
(5.5) + 2 = (d - 2) + 2
7.5 = d

Now if instead of adding 2, you subtract 2, the problem becomes:
5.5 = d - 2
(5.5) - 2 = (d - 2) - 2
3.5 = d - 4

This doesn't solve your problem, it just makes the value needed to be canceled out larger.

Thus d = 7.5 and you will add because the value is negative.

Akimi4 [234]4 years ago

5 0

Since 2 is negative, you would want to add the 2 to each side. You want to cancel out the number so you leave d by itself, so it'll always be opposite of whatever the sign is.
5.5= -2 + d
+2 +2
7.5 = d
Hope this helped!

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Answer:

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$p_v =P(Z>3.37)=1-P(Z$

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Step-by-step explanation:

Data given

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1 330 33 8

2 310 31 7

$\bar X_{1}=33$ represent the mean for sample 1

$\bar X_{2}=31$ represent the mean for sample 2

$\sigma_{1}=8$ represent the population standard deviation for 1

$\sigma_{2}=7$ represent the population standard deviation for 2

$n_{1}=330$ sample size for the group 1

$n_{2}=310$ sample size for the group 2

$\alpha$ Significance level provided

z would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean for Campus 1 is higher than the mean for Campus 2, the system of hypothesis would be:

Null hypothesis: $\mu_{1}-\mu_{2}\leq 0$

Alternative hypothesis: $\mu_{1} - \mu_{2}> 0$

We have the population standard deviation's, and the sample sizes are large enough we can apply a z test to compare means, and the statistic is given by:

$z=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (1)

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

With the info given we can replace in formula (1) like this:

$z=\frac{(33-31)-0}{\sqrt{\frac{8^2}{330}+\frac{7^2}{310}}}}=3.37$

P value

Since is a one right tailed test the p value would be:

$p_v =P(Z>3.37)=1-P(Z$

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