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AnnyKZ [126]
3 years ago
8

Solve for m<1 and m<2.

Mathematics
1 answer:
stira [4]3 years ago
8 0

Answer:

M<2 =61* M< = 61

Step-by-step explanation:

You know angle 4 is 119* and the straight line is 180* so if u do 180 - 119 it is 61 , therefore 1 ,2,and 3 are 61*

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What is the value of p-12=14
Trava [24]

Answer:

the value of p? its 26 because 26-12=14

Step-by-step explanation:

8 0
3 years ago
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X+8y<br> units<br> 8x -x+5y units<br> what’s the perimeter?
pochemuha

Answer:

P = 16x+26y units

Step-by-step explanation:

Let the length is (x+8y) units

Width = (8x-x+5y) units

We need to find the perimeter. We know that the perimeter is equal to the sum of all sides.

P = 2[(x+8y) + (8x-x+5y)]

= 2[(8x+8y+5y)]

= 2[8x+13y]

= 16x+26y

Hence, the required perimeter is 16x+26y.

3 0
3 years ago
What do both of these functions have in common? f(x)=5e^x+5 g(x)=0.5(x-5)^2-5
kondor19780726 [428]

Answer:

They have the same vertical shift. -Apex

Step-by-step explanation:

3 0
3 years ago
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Experian would like to test the hypothesis that the average credit score for an adult in Virginia is different from the average
aliya0001 [1]

Answer:

a. We fail to reject the hypothesis that the average credit score for an adult in Virginia is different from the average credit score for an adult in North Carolina at the significance level of 0.05.

b. The 95% confidence interval for the true difference of means is -2.2468 and 36.2468. There is a probability of 95% that the true difference of means \mu_{1}-\mu_{2} is between -2.2468 and 36.2468. This confidence interval contains the number 0, this is consistent with the results we got in a. Because we fail to reject the hypothesis that the average credit score for an adult in Virginia is different from the average credit score for an adult in North Carolina at the significance level of 0.05.

Step-by-step explanation:

Let \mu_{1}-\mu_{2} be the true difference between the average credit score for an adult in Virginia and the average credit score for an adult in North Carolina. We have the large sample sizes n_{1} = 40 and n_{2} = 35, the unbiased point estimate for \mu_{1}-\mu_{2} is \bar{x}_{1} - \bar{x}_{2}, i.e., 699-682 = 17.

The standard error is given by \sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}, i.e.,

\sqrt{\frac{(44)^{2}}{40}+\frac{(41)^{2}}{35}} = 9.8198.

a. We want to test H_{0}: \mu_{1}-\mu_{2} = 0 vs H_{1}: \mu_{1}-\mu_{2} \neq 0 (two-tailed alternative). The rejection region is given by RR = {z | z < -1.96 or z > 1.96} where -1.96 and 1.96 are the 2.5th and 97.5th quantiles of the standard normal distribution respectively. The test statistic is Z = \frac{\bar{x}_{1} - \bar{x}_{2}-0}{\sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}} and the observed value is z_{0} = \frac{17}{9.8198} = 1.7312. Because 1.7312 does not fall inside RR, we fail to reject the null hypothesis.

b. The endpoints for a 95% confidence interval for \mu_{1}-\mu_{2} is given by 17\pm (z_{0.05/2})9.8198, i.e., 17\pm (z_{0.025})9.8198 where z_{0.025} is the 2.5th quantile of the standard normal distribution, i.e., -1.96, so, we have 17-(1.96)(9.8198) and 17+(1.96)(9.8198), i.e., -2.2468 and 36.2468. There is a probability of 95% that the true difference of means \mu_{1}-\mu_{2} is between -2.2468 and 36.2468. This confidence interval contains the number 0, this is consistent with the results we got in a. Because we fail to reject the hypothesis that the average credit score for an adult in Virginia is different from the average credit score for an adult in North Carolina at the significance level of 0.05.

3 0
3 years ago
Just need to know the answer thanks
enyata [817]
A) x=3   x=2
b) x=-8   x=-2

6 0
3 years ago
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