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3 years ago

6

write the ordered pair for each point. PLS! I will give 20 points did correct same goes with brainliest. DO NOT TYPE THAT YOU DO

N'T KNOW OR I WILL FLAG, REPORT AND GET MAD!

1 answer:

patriot [66]3 years ago

4 0

Answer:

M=(0.75,1.25)

H=(-1.5,-0.5)

D=(1,-1.75)

Step-by-step explanation:

Hope this helps.

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Which expression represents the correct form of the partial fraction decomposition of 19x + 20 answer asap

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3 years ago

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you start out driving from your house to an old friends house in a different town, which is 280 miles away. after an hour and a

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2 years ago

A student at a four-year college claims that mean enrollment at four-year colleges is higher than at two-year colleges in the Un

WINSTONCH [101]

Answer:

Part 1: The statistic

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

And the degrees of freedom are given by $df=n_1 +n_2 -2=35+35-2=68$

Replacing we got

$t=\frac{(5135-4436)-0}{\sqrt{\frac{783^2}{35}+\frac{553^2}{35}}}}=4.31$

Part 2: P value

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

$p_v =P(t_{68}>4.31)=0.000022 \approx 0.00002$

Comparing the p value we see that is lower compared to the significance level of 0.01 so then we can reject the null hypothesis and we can conclude that the mean for the four year college is significantly higher than the mean for the two year college and then the claim makes sense

Step-by-step explanation:

Data given

$\bar X_{1}=5135$ represent the mean for four year college

$\bar X_{2}=4436$ represent the mean for two year college

$s_{1}=783$ represent the sample standard deviation for four year college

$s_{2}=553$ represent the sample standard deviation two year college

$n_{1}=35$ sample size for the group four year college

$n_{2}=35$ sample size for the group two year college

$\alpha=0.01$ Significance level provided

t would represent the statistic (variable of interest)

System of hypothesis

We need to conduct a hypothesis in order to check if the mean enrollment at four-year colleges is higher than at two-year colleges in the United States , the system of hypothesis would be:

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

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

We can assume that the normal distribution is assumed since we have a large sample size for each case n>30. So then the sample mean can be assumed as normally distributed.

Part 1: The statistic

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

And the degrees of freedom are given by $df=n_1 +n_2 -2=35+35-2=68$

Replacing we got

$t=\frac{(5135-4436)-0}{\sqrt{\frac{783^2}{35}+\frac{553^2}{35}}}}=4.31$

Part 2: P value

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

$p_v =P(t_{68}>4.31)=0.000022$

Comparing the p value we see that is lower compared to the significance level of 0.01 so then we can reject the null hypothesis and we can conclude that the mean for the four year college is significantly higher than the mean for the two year college and then the claim makes sense

6 0

3 years ago

How do you get the numbers: 7,6,5, and 3 to equal 75. only using eachnumber once?? thanks,

Margaret [11]

You might have to use parentheses to get 75...
The answer is
6(7+5) + 3

Need an explanation?

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3 years ago

Helppppppppppping me please

Gala2k [10]

Answer:

<h2>8. D) 2x + 8</h2><h2>9. C) 6x + 32 = 158</h2><h2>10. 50 ft</h2>

Step-by-step explanation:

Look at the picture.

9. The perimeter of the rectangle l × w:

P = 2l + 2w

Substitute l = 2x + 8 and w = x + 8:

P = 2(2x + 8) + 2(x + 8) <em>use the distributive property</em>

P = (2)(2x) + (2)(8) + (2)(x) + (2)(8)

P = 4x + 16 + 2x + 16 <em>combine like terms</em>

P = (4x + 2x) + (16 + 16)

P = 6x + 32

10. Solve the equation:

6x + 32 = 158 <em>subtract 32 from both sides</em>

6x = 126 <em>divide both sides by 6</em>

x = 21

Put the value of x to the expression 2x + 8:

2(21) + 8 = 42 + 8 = 50

4 0

3 years ago