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vladimir1956 [14]
3 years ago
13

What is the midpoint of the segment shown below?

Mathematics
1 answer:
Papessa [141]3 years ago
4 0
B bro it b just trust I promise it b
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Please hurry I will mark you brainliest <br> Find the surface area
erastova [34]

Triangle area = 8 x 3/2

= 12

12 x 2 = 24cm^2 for 2 triangles

Height of rectangle must be found, so use pythag on triangle

a^2+ b^2 = c^2

16 + 9 =c^2

C=5

A= 5 x 12

= 60

60 x 3 triangles in total is 180cm^2

180cm^2 + 24cm^2 = 204cm^2

Ans: 204cm^2

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Solve these liner inequalities 2(x+1)&gt;3(x-8)
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The answer for that problem is X<26
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Is f(x) = sec x concave up or concave down at x = 17\pi/4<br>​
inessss [21]

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x^xtfr

Step-by-step explanation:

6 0
3 years ago
Suppose that scores on the mathematics part of a test for eighth-grade students follow a Normal distribution with standard devia
riadik2000 [5.3K]

Answer:

We need an SRS of scores of at least 153.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.9}{2} = 0.05

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.05 = 0.95, so z = 1.645

Now, find M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

How large an SRS of scores must you choose?

This is at least n, in which n is found when M = 20, \sigma = 150. So

M = z*\frac{\sigma}{\sqrt{n}}

20 = 1.645*\frac{150}{\sqrt{n}}

20\sqrt{n} = 1.645*150

\sqrt{n} = \frac{1.645*150}{20}

\sqrt{n} = 12.3375

(\sqrt{n})^{2} = (12.3375)^{2}

n = 152.2

Rounding to the next whole number, 153

We need an SRS of scores of at least 153.

8 0
3 years ago
if a committee of 4 is to be chosen at random from the class of 11 students what is the probability of any particular commuter b
Darina [25.2K]

i think the answer is 4/11 because the 4 people has a 1%  chance out of the 11 people

7 0
3 years ago
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