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DochEvi [55]
3 years ago
9

A circle has a radius of 3 cm. What is its: a. Diameter? b. Area? c. Circumference?

Mathematics
2 answers:
larisa86 [58]3 years ago
7 0

Answer:

Its B

Step-by-step explanation:

Virty [35]3 years ago
7 0
B- area is the correct answer
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Round 146,273,011 to the nearest hundred thousand.<br><br> i need help
loris [4]

Answer:

146,300,000

Step-by-step explanation:

146,273,011

<u>The hundred thousand place is 100,000</u>

<u />

<u>Step 1:  Round</u>

146,273,011

<em>146,300,000</em>

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Answer:  146,300,000

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3 years ago
FIND THE VALUE OF X AND Y.(ordered pairs)<br><br>(x+y,0)=(8,x-y)​
Aliun [14]

Step-by-step explanation:

We know that...

x+y (from the first pair) corresponds to 8(in the second pair)..

0 (from the first pair) corresponds to x-y (in the second pair)..

Hence,

x + y = 8 \\ x - y = 0

By simply adding the two equations we get,

2x = 8 \\ x  = 4

Hence,

By substituting x=4, we get that y=4

Hence,

x=y=4

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3 years ago
An electronics company just finished designing a new tablet computer and is interested in estimating its battery-life. A random
Alika [10]

Answer:

6 \pm 2.861(0.3354)

Step-by-step explanation:

We have the standard deviation for the sample, so we use the t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 20 - 1 = 19

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 19 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.99}{2} = 0.995. So we have T = 2.861.

The margin of error is:

M = T\frac{s}{\sqrt{n}} = 2.861\frac{1.5}{\sqrt{20}} = 2.861(0.3354)

In which s is the standard deviation of the sample and n is the size of the sample.

The format of the confidence interval is:

S_{M} \pm M

In which S_{M} is the sample mean

So

6 \pm 2.861(0.3354)

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natita [175]
THE ANSWER IS 12(sorry for the caps)
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I don’t really understand this stuff. Anyone?
ycow [4]
MA is the bisector of RS
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3 years ago
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