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irakobra [83]
3 years ago
10

Alpha JKL has vertices J(-3,7), K(2,8), and L(-3, 12). Find the perimeter of AJKL to the nearest tenth.

Mathematics
1 answer:
Crazy boy [7]3 years ago
3 0

Answer:

Alpha JKL has vertices J(-3,7), K(2,8), and L(-3, 12). Find ...brainly.com › Mathematics › High School

20 hours ago — Alpha JKL has vertices J(-3,7), K(2,8), and L(-3, 12). Find the perimeter of AJKL to the nearest tenth. - 18900060.

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

Use rule that a straight line is 180 and a right angle is 90

8 0
2 years ago
What is the median of the data set?
cluponka [151]

Answer: 85.5

Step-by-step explanation:

There are 22 data points in the data set.

We can find the median by looking at the 11th and 12th scores, and averaging them.

The 11th score is 85

The 12th score is 86

The average of the two is 85.5

Hence the answer is 85.5

3 0
3 years ago
Write the equation of the line that passes through the given point and has the given slope.
Vilka [71]
The answer is y=3x-10
6 0
2 years ago
Find the coordinates of the midpoint of the segment whose endpoints are H(5, 13) and K(7, 5). (12, 18) (9, 7) (2, 8) (6, 9)
____ [38]

Answer: D. (6, 9)

<u>Step-by-step explanation:</u>

Midpoint is the "average" of the x's and y's:

Given: (5, 13) and (7, 5)

Midpoint: (\dfrac{5+7}{2},\dfrac{13+5}{2})

             = (\dfrac{12}{2},\dfrac{18}{2})

              = (6, 9)

4 0
3 years ago
A genetic experiment involving peas yielded one sample of offspring consisting of 409 green peas and 171 yellow peas. Use a 0.05
grigory [225]
The usual expectation for this kind of experiment is that the peas would yield green and yellow peas in a 3:1 ratio, or 75% green to 25% yellow. So your null hypothesis is that the proportion of yellow peas is p=0.25.

You're testing the claim that 26% of the offspring will be yellow, which means the alternative hypothesis is that the proportion of yellow peas is actually 0.26, or more generally that the expected proportion is greater than 0.25, or p>0.25.

The test statistic in this case will be

Z=\dfrac{\hat p-p_0}{\sqrt{\dfrac{p_0(1-p_0)}n}}

where p_0 is the proportion assumed under the null hypothesis, \hat p is the measured proportion, and n is the sample size. You have p_0=0.25, \hat p=\dfrac{171}{171+409}\approx0.29, and n=409+171=580, so the test statistic is

Z=\dfrac{0.29-0.25}{\sqrt{\dfrac{0.25\times0.75}{580}}}\approx2.2247

Because you're testing p>p_0, this is a right-tailed test, so the P-value is

\mathbb P(Z>2.2247)\approx0.0131

The critical value for a right-tail test at a 0.05 significance level is Z_\alpha\approx1.6449, which means the rejection region is any test statistic that is larger than this critical value. Since Z>Z_\alpha in this case, we reject the null hypothesis.

So the conclusion for this test is that the sample proportion is indeed statistically significantly different from the proportion suggested by the null hypothesis.
8 0
3 years ago
Read 2 more answers
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