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White raven [17]
3 years ago
11

CORRECT ANSWERS WILL GET 30 POINTS!!! PLEASE HELP

Mathematics
2 answers:
irakobra [83]3 years ago
6 0

Answer:

B

sqrt of 1000= 31.6m

A=pi r2

1000pi= pi r2 (pi cancels)

1000=r2 (take square root of both sides)

r= 31.6m

Step-by-step explanation:

Vlad [161]3 years ago
4 0
R=31.6m is the answer
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If 60% of a KNOWN NUMBER is 180, what is the 40% of that UNKNOWN NUMBER?
Lana71 [14]

Step-by-step explanation:

I think you mean you want to find out 40% of a number,, when 60% of it is 180, so I'll answer that:

60% = 180

÷6

10% = 30

x4

40% = 120

(if you want to find 100%, multiply 30 by 10)

Hopefully this helps :)

4 0
2 years ago
Help me please and the last option is 9cm on the left and 12 on bottom
Rasek [7]
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3 years ago
In selecting a sulfur concrete for roadway construction in regions that experience heavy frost, it is important that the chosen
Lesechka [4]

Step-by-step explanation:

Given - In selecting a sulfur concrete for roadway construction in regions that experience heavy frost, it is important that the chosen concrete has a low value of thermal conductivity in order to minimize subsequent damage due to changing temperatures. Suppose two types of concrete, a graded aggregate and a no-fines aggregate, are being considered for a certain road. The table below summarizes data on thermal conductivity from an experiment carried out to compare the two types of concrete.

Type            ni                xi                  si

Graded       42           0.486            0.187

No-fines      42           0.359           0.158

To find - a. Formulate the above in terms of a hypothesis testing problem.

b. Give the test statistic and its reference distribution (under the null hypothesis).

c. Report the p-value of the test statistic and use it to assess the evidence that this sample provides on the scientific question of difference in mean conductivity of the two materials at the 5% level of significance.

Proof -

a.)

Hypothesis testing problem :

H0 : There is significant difference between mean conductivity for the graded concrete and mean conductivity for the no fines concrete.

H1 : There is no significant difference between mean conductivity for the graded concrete and mean conductivity for the no fines concrete.

b)

Test statistic :

Z = \frac{x_{1} - x_{2} }{\sqrt{\frac{s_{1} ^{2} }{n_{1}}  + \frac{s_{2}^{2} }{n_{2}} } }

Z = \frac{0.486 - 0.359 }{\sqrt{\frac{0.03496 }{42}  + \frac{0.02496 }{42} } }

Z = \frac{0.127 }{\sqrt{0.001468}}

Z = \frac{0.127 }{0.0377}

⇒Z(cal) = 3.3687

Z(tab) = 1.96

As Z (cal) > Z(tab)

So, we reject H0 at 5% Level of significance

p-value = 0.99962

Hence

There is significant difference in mean conductivity at the two materials.

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