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liberstina [14]
2 years ago
10

20 for this question needing help

Mathematics
1 answer:
Radda [10]2 years ago
7 0

Step-by-step explanation:

option C is the answer.

hope it helps

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Sammy has x flavors of candies with which to make goody bags for Frank's birthday party. Sammy tosses out y flavors, because he
harkovskaia [24]

Answer:

^{(x-y)}C_{10}=\frac{(x-y)!}{10! \times (x-y-10)!}

Step-by-step explanation:

Total flavors Sammy initially had = x

Number of flavors Sammy throw away = y

After throwing away y flavors, the number of flavors Sammy will be left with = x - y

He needs to make 10-flavor bags from these (x - y) flavors. In order words he needs to chose 10 flavors for each bag from(x - y) flavors. The order of selection is not important here, so this is a problem of combinations. Also since we have to make selections or small groups, this also indicates that we have to use combinations.

So we need to make combinations of 10 flavors from a total of (x - y) flavors. This can be represented as ^{(x-y)}C_{10}

The formula for combinations is:

^{n}C_{r}=\frac{n!}{r!(n-r)!}

Using the values in this formula, we get:

^{(x-y)}C_{10}=\frac{(x-y)!}{10! \times (x-y-10)!}

7 0
3 years ago
What is the value of f(2)?
Lunna [17]

Question:

What is the equation?

6 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
2 years ago
What is 10 × 3 1/6 in it simplest form
Ray Of Light [21]

Answer:

Step-by-step explanation:10x3 1/6=30 10/6=31 2/6=31 1/3

3 0
3 years ago
8(1/2n+3/4)=70 solve for n
nasty-shy [4]

Answer:

n = 16

Step-by-step explanation:

8 0
3 years ago
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