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3 years ago

ZE and ZF are vertical angles with mZE = 8x + 8 and mZF = 2x +38. What is the value of x?

Mathematics

2 answers:

Dovator [93]3 years ago

7 0

Answer:

Since ZE and ZF are vertical angles, they are congruent.

8x+8 = 2x+38

8 = 2x-8x+38

8 = -6x+38

8-38 = -6x

-30 = -6x

x = -30/-6

x = 5

mash [69]3 years ago

3 0

Answer:

x = 5

Step-by-step explanation:

Vertical angles are congruent, so ...

∠E = ∠F

8x +8 = 2x +38

6x = 30 . . . . . . . . . subtract (2x+8)

x = 5 . . . . . . . . . . . divide by 6

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In a set of 25 aluminum castings, four castings are defective (D), and the remaining twenty-one are good (G). A quality control

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Answer:

The sample space for selecting the group to test contains 2,300 elementary events.

Step-by-step explanation:

There are a total of N = 25 aluminum castings.

Of these 25 aluminum castings, n₁ = 4 castings are defective (D) and n₂ = 21 are good (G).

It is provided that a quality control inspector randomly selects three of the twenty-five castings without replacement to test.

In mathematics, the procedure to select k items from n distinct items, without replacement, is known as combinations.

The formula to compute the combinations of k items from n is given by the formula:

${n\choose k}=\frac{n!}{k!(n-k)!}$

Compute the number of samples that are possible as follows:

${25\choose 3}=\frac{25!}{3!\times (25-3)!}$

$=\frac{25\times 24\times 23\times 22!}{3!\times 22!}\\\\=\frac{25\times 24\times 23}{3\times 2\times 1}\\\\=2300$

The sample space for selecting the group to test contains 2,300 elementary events.

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3 years ago

You randomly draw two cards from a standard deck of playing cards without replacement.what is the probability that both of the c

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There are 12 face cards in a deck, and one you draw the first face card there will be 11 left so:
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Assume that a policyholder is four times more likely to file exactly two claims as to file exactly three claims. Assume also tha

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Answer:

1.3125

Step-by-step explanation:

Given that our random variable $X$ follows a Poisson distribution $P(X=k)=\frac{\lambda^k e^-^p}{k!} \ \ \ \ \ \ \ \ p=\lambda$

Evaluate the formula at $k=2,3:$

$P(X=2)=0.5\lambda^2e^{-\lambda}\\\\P(X=3)=\frac{1}{6}\lambda^2e^{-\lambda}$

#since

$4P(X=2)=P(X=3);\\\\0.5\lambda^2e^{-\lambda}=4\times\frac{1}{6}\lambda^3e^{-\lambda}\\\\0.5\lambda^2=\frac{2}{3}\lambda^3\\\\0.75=\lambda$

The mean and variance of the Poisson distributed random variable is equal to $\lambda$ :

$\mu=\lambda=0.75\\\sigma ^2=\lambda=0.75$

#By property variance:

$\sigma ^2=V(X)=E(X^2)-(E(X))^2=E(X^2)\\\\E(X^2)=\sigma^2+\mu^2=0.75+0.75^2=1.3125$

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3 years ago

Evaluate the function at the indicated values. f(x)= 6x^2+ 5x -12

seropon [69]

Step-by-step explanation:

We need to find the value of the function at the indicated value.

$f(x)= 6x^2+ 5x -12$

Indicated values are : f(-7), f(0), f(6), f(7)

To find f(-7), put x = -7 in the given function

$f(-7)= 6(-7)^2+ 5(-7) -12=247$

To find f(0), put x = 0 in the given function

$f(0)= 6(0)^2+ 5(0) -12=-12$

To find f(6), put x = 6 in the given function

$f(6)= 6(6)^2+ 5(6) -12=234$

To find f(7), put x = 7 in the given function

$f(7)= 6(7)^2+ 5(7) -12=317$

Hence, this is the required solution.

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