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daser333 [38]
3 years ago
9

given that measurement is in centimeters, find the circumference of the circle of the nearest tenth. (use 3.14 for pi

Mathematics
2 answers:
Elza [17]3 years ago
6 0

Answer: <em>31.4cm is the correct answer</em>

Step-by-step explanation:

<em>31.4 cm </em>

<em> </em>

<em>d = 2r = (2)(5) = 10 </em>

<em> </em>

<em>C = πd = 10π = 31.4</em>

Ivahew [28]3 years ago
4 0

Answer:

C = 18.8 cm

Step-by-step explanation:

The diameter is 6 cm

To find the circumference, we use

C = pi*d

C = 3.14 *6

C =18.84 cm

Rounding to the nearest tenth

C = 18.8 cm

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Which of the following information about the graph of the relationship is given
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Answer:

there is no gragh

Step-by-step explanation:

6 0
3 years ago
Let's try an application problem!
mamaluj [8]

Answer: 5 ft

Step-by-step explanation:

Area of a rectangle (which is the shape of a regular flag) = L* W.

W = Area/L = 9.5/1.9 = 5ft.

8 0
3 years ago
A quality control expert at Glotech computers wants to test their new monitors. The production manager claims they have a mean l
tankabanditka [31]

Answer:

The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors

P(X⁻ ≥ 96.3) = 0.0087

Step-by-step explanation:

<u><em>Step(i):-</em></u>

Given that the mean of the Population = 95

Given that the standard deviation of the Population = 5

Let 'X' be the random variable in a normal distribution

Let X⁻ = 96.3

Given that the size 'n' = 84 monitors

<u><em>Step(ii):-</em></u>

<u><em>The Empirical rule</em></u>

         Z = \frac{x^{-} -mean}{\frac{S.D}{\sqrt{n} } }

        Z = \frac{96.3 - 95}{\frac{5}{\sqrt{84} } }

       Z = 2.383

The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors

    P(X⁻ ≥ 96.3) = P(Z≥2.383)

                      =  1- P( Z<2.383)

                      =  1-( 0.5 -+A(2.38))

                      = 0.5 - A(2.38)

                     = 0.5 -0.4913

                    = 0.0087

<u><em>Final answer:-</em></u>

The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors

P(X⁻ ≥ 96.3) = 0.0087

3 0
3 years ago
How do the values in Pascal’s triangle connect to the coefficients?
damaskus [11]

Explanation:

Each row in Pascal's triangle is a listing of the values of nCk = n!/(k!(n-k)!) for some fixed n and k in the range 0 to n. nCk is <em>the number of combinations of n things taken k at a time</em>.

If you consider what happens when you multiply out the product (a +b)^n, you can see where the coefficients nCk come from. For example, consider the cube ...

  (a +b)^3 = (a +b)(a +b)(a +b)

The highest-degree "a" term will be a^3, the result of multiplying together the first terms of each of the binomials.

The term a^b will have a coefficient that reflects the sum of all the ways you can get a^b by multiplying different combinations of the terms. Here they are ...

  • (a +_)(a +_)(_ +b) = a·a·b = a^2b
  • (a +_)(_ +b)(a +_) = a·b·a = a^2b
  • (_ +b)(a +_)(a +_) = b·a·a = a^2b

Adding these three products together gives 3a^2b, the second term of the expansion.

For this cubic, the third term of the expansion is the sum of the ways you can get ab^2. It is essentially what is shown above, but with "a" and "b" swapped. Hence, there are 3 combinations, and the total is 3ab^2.

Of course, there is only one way to get b^3.

So the expansion of the cube (a+b)^3 is ...

  (a +b)^3 = a^3 + 3a^2b +3ab^2 +b^3 . . . . . with coefficients 1, 3, 3, 1 matching the 4th row of Pascal's triangle.

__

In short, the values in Pascal's triangle are the values of the number of combinations of n things taken k at a time. The coefficients of a binomial expansion are also the number of combinations of n things taken k at a time. Each term of the expansion of (a+b)^n is of the form (nCk)·a^(n-k)·b^k for k =0 to n.

6 0
3 years ago
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