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inysia [295]
3 years ago
9

Please help the following questions

Mathematics
1 answer:
insens350 [35]3 years ago
6 0

Answer:

<em>Option D</em>

Step-by-step explanation:

Assume that these polygons were similar. You could tell that if they were, ABCD ~ EFGH, and is already noted by the answer choices. We can conclude that AB ~ EF, CD ~ GH and so on. To prove that these shapes are similar, we must form a like proportionality among these side;

EF / AB = GH / CD\\5 / 8 = 4 / 2.5,\\0.625 \neq 1.6\\\\Conclusion - Option D

As the proportions we formed were to equal to one another,<em> the solution must be option D!</em>

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The two pictures are the questions and question five is what i need help with.
Mama L [17]

Answer:yes they do

Step-by-step explanation:

5 0
2 years ago
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Help on questions 37 and 38 please!
jeka94

Answer:

Step-by-step explanation:

38.

M=\frac{1}{2} (x_{1}+x_{2})\\2M=x_{1}+x_{2}\\x_{1}=2M-x_{2}

37.

S=\frac{ab^2}{3} \\3S=ab^2\\b^2=\frac{3S}{a} \\b=\pm\sqrt{\frac{3S}{a} }

6 0
3 years ago
The table shows the height, in centimeters, that a weight bouncing from a spring would achieve if there were no friction, for a
Paha777 [63]
Hello! For ease of calculations, we can identify the time it took for the weight to bounce back to the other direction, then the other, and then back to its original position by looking at the time it took for the weight to change from 0 to 25 to 0 to -25 then back to 0. This is one whole cycle of the weight.

By the time the weight first reached zero, 1.5 seconds has passed. By the third time it got to zero again, 7.5 seconds has passed. Therefore, one whole cycle of the weight is 7.5-1.5 = 6.0 seconds.

ANSWER: One whole cycle of the weight took 6 seconds.
4 0
3 years ago
2x + y = 4<br> y = 32 – 1<br> Is (1, 2) a solution to both equations?
Digiron [165]

Answer:

No

Step-by-step explanation:

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6 0
3 years ago
An average computer mouse inspector can inspect 50 mice per hour. The 48 computer mice inspectors at a particular factory can on
Ronch [10]

Answer:  C. Yes, because -2.77 falls in the critical region .

Step-by-step explanation:

Let \mu be the population mean .

As per given , we have

H_0:\mu=50\\\\ H_a: \mu

Since the alternative hypothesis is left-tailed and population standard deviation is not given , so we need to perform a left-tailed t-test.

Test statistic : t=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}

Also, it is given that ,

n= 48

\overline{x}=46

s= 10

t=\dfrac{46-50}{\dfrac{10}{\sqrt{48}}}=\dfrac{-4}{\dfrac{10}{6.93}}\\\\=\dfrac{-4}{1.443}\approx-2.77

Degree of freedom = df = n-1= 47

Using t-distribution , we have

Critical value =t_{\alpha,df}=t_{0.025,47}=2.0117

Since, the absolute t-value (|-2.77|=2.77) is greater than the critical value.

So , we reject the null hypothesis.

i.e. -2.77 falls in the critical region.

[Critical region is the region of values that associates with the rejection of the null hypothesis at a given probability level.]

Conclusion : We have sufficient evidence to support the claim that these inspectors are slower than average.

Hence, the correct answer is C. Yes, because -2.77 falls in the critical region

4 0
3 years ago
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