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2 years ago

Thank you guys fir the help

Mathematics

1 answer:

KiRa [710]2 years ago

7 0

9514 1404 393

Answer:

Step-by-step explanation:

The function f(x) is required in the numerator, eliminating choices C and D.

The restriction is that function g cannot be zero, so we cannot have ...

3x +2 = 0

3x = -2

x = -2/3 . . . . . eliminates choice B; confirms choice A

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A supplier delivers an order for 20 electric toothbrushes to a store. By accident, three of the electric toothbrushes are defect

krek1111 [17]

Answer:

The probability that the first two electric toothbrushes sold are defective is 0.016.

Step-by-step explanation:

The probability of an event, say E occurring is:

$P (E)=\frac{n(E)}{N}$

Here,

n (E) = favorable outcomes

N = total number of outcomes

Let X = number of defective electric toothbrushes sold.

The number of electric toothbrushes that were delivered to a store is, n = 20.

Number of defective electric toothbrushes is, x = 3.

The number of ways to select two toothbrushes to sell from the 20 toothbrushes is:

${20\choose 2}=\frac{20!}{2!(20-2)!}=\frac{20!}{2!\times 18!}=\frac{20\times 19\times 18!}{2!\times 18!}=190$

The number of ways to select two defective toothbrushes to sell from the 3 defective toothbrushes is:

${3\choose 2}=\frac{3!}{2!(3-2)!}=\frac{3!}{2!\times 1!}=3$

Compute the probability that the first two electric toothbrushes sold are defective as follows:

P (Selling 2 defective toothbrushes) = Favorable outcomes ÷ Total no. of outcomes

$=\frac{3}{190}\\$

$=0.01579\\\approx0.016$

Thus, the probability that the first two electric toothbrushes sold are defective is 0.016.

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The sum will be:

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Answer:

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