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2 years ago

11

A circle representing a pool is graphed with a center at the origin. Grant enters the pool at point A and swims over to a friend

who is located at point B.

1 answer:

Anit [1.1K]2 years ago

4 0

Grant would swim three the middles of the pool (just doing this for the points)

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The length of one of the diagonals of a rhombus is 6, and the perimeter is 20. What is the length of the other diagonal?

Lelechka [254]

Answer:

4

Step-by-step explanation:

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3 years ago

(-5b2 – 8b) -(-9b3– 5b2 – 8b)

Dafna11 [192]

Answer:

9b³

Step-by-step explanation:

-5b²-8b+9b³+5b²+8b

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4 years ago

DEPERATE PLZ HELP IM ON A TIMED TEST

Crank

Answer:

y = 6x - 2

Step-by-step explanation:

Let the equation of the line be y = mx + c

m = (10 - (-8))/(2 - (-1)) = 6

sub (2, 10):

10 = 6(2) + c

c = -2

therefore, the equation of the line is y = 6x - 2

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3 years ago

Please Help will award brainliest

EastWind [94]

Answer:7

Step-by-step explanation:

Remember,

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Begin by 3 / 1/3 = 1

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3 years ago

A factory tests 100 light bulbs for defects. The probability that a bulb is defective is 0.02. The occurrences of defects among

Kobotan [32]

Answer:

The probability that exactly two are defective given that the number of defective bulbs is two or fewer is 0.4040.

Step-by-step explanation:

Let X be the number of defective bulbs.

$X\sim B(n,p)$

Where n is sample size and p is probability of success.

According to the given information,

$n=100$

$p=0.02$

$q=1-p=1-0.02=0.98$

According to binomial distribution the probability of exactly r success from n is

$P(X=r)=^nC_rp^rq^{n-r}$

The probability that exactly two are defective is

$P(X=2)=^{100}C_2(0.02)^2(0.98)^{98}\approx 0.2724$

The probability that the number of defective bulbs is two or fewer is

$P(X\leq 2)=P(X=0)+P(X=1)+P(X=2)$

$P(X\leq 2)=^{100}C_0(0.02)^0(0.98)^{100}+^{100}C_1(0.02)^1(0.98)^{99}+^{100}C_2(0.02)^2(0.98)^{98}$

$P(X\leq 2)=0.1326+0.2707+0.2734$

$P(X\leq 2)=0.6767$

We have to find the probability that exactly two are defective given that the number of defective bulbs is two or fewer.

$P(x=2|x\leq 2)$

According to the conditional probability

$P(\frac{A}{B})=\frac{P(A\cap B)}{P(B)}$

$P(x=2|x\leq 2)=\frac{P(x=2\cap x\leq 2)}{P(x\leq 2)}$

$P(x=2|x\leq 2)=\frac{P(x=2)}{P(x\leq 2)}$

$P(x=2|x\leq 2)=\frac{0.2734}{0.6767}=0.4040$

Therefore the probability that exactly two are defective given that the number of defective bulbs is two or fewer is 0.4040.

6 0

4 years ago