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Verdich [7]
3 years ago
14

Find the approximate value of the circumference of a circle with the given radius. Use = 3.14. Round your answer to the nearest

hundredth.
5.1 cm
Mathematics
1 answer:
blsea [12.9K]3 years ago
7 0
The circumference of a circle:
C = 2 r π
If r = 5.1 cm:
C = 2 · 5.1 · 3.14 = 32.028 cm ≈ 32.03 cm
Answer:
The approximate value of the circumference is 32.03 cm ( rounded to the nearest hundredth ).
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John is expanding these two brackets.
Elan Coil [88]

Answer: Um, the answer is B, and C?

Step-by-step explanation:

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In Aunt Melly’s attics there are spiders and ants, the total of 152 legs and 22 heads. How many spiders and how many ants?
Karolina [17]
152+22 =372 i think it is that answer
6 0
3 years ago
What are the dimensions for this triangle?
REY [17]

Answer:

2 and 2

Step-by-step explanation:

Parallel sides of rectangles are congruent, and because the other side of both rectangles is 2, the sides of the triangle are 2 because they share the side with the rectangle.

4 0
4 years ago
Write an algebraic rule to describe the translation c (5,-4) c'(-2,1)
Sauron [17]
We can solve this by using the formula:
(x, y) (x + a, y + b) = (5,-4) (-2,1)

So, plugging in the values and solving for a and b,
5 + a = -2
a = -8

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b = 5

Therefore, the translation is
(x,y) (x - 8, y +5)
5 0
3 years ago
A survey was conducted that asked 1003 people how many books they had read in the past year. Results indicated that x overbar eq
Sergio [31]

Answer:

The 99% confidence interval would be given (11.448;14.152).

Step-by-step explanation:

1) Important concepts and notation

A confidence interval for a mean "gives us a range of plausible values for the population mean. If a confidence interval does not include a particular value, we can say that it is not likely that the particular value is the true population mean"

s=16.6 represent the sample deviation

\bar X=12.8 represent the sample mean

n =1003 is the sample size selected

Confidence =99% or 0.99

\alpha=1-0.99=0.01 represent the significance level.

2) Solution to the problem

The confidence interval for the mean would be given by this formula

\bar X \pm z_{\alpha/2} \frac{s}{\sqrt{n}}

We can use a z quantile instead of t since the sample size is large enough.

For the 99% confidence interval the value of \alpha=1-0.99=0.01 and \alpha/2=0.005, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=2.58

And replacing into the confidence interval formula we got:

12.8 - 2.58 \frac{16.6}{\sqrt{1003}}=11.448

12.8 + 2.58 \frac{16.6}{\sqrt{1003}} =14.152

And the 99% confidence interval would be given (11.448;14.152).

We are confident that about 11 to 14 are the number of books that the people had read on the last year on average, at 1% of significance.

3 0
3 years ago
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