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3 years ago

1f (-2, y) lies on the graph of y = 3, then y =

Mathematics

1 answer:

kkurt [141]3 years ago

7 0

Answer:

y = 3

Step-by-step explanation:

the y-value of any ordered pair that lies on the line 'y = 3' will be equal to 3.

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Simplify:

adelina 88 [10]

Answer:

The answer is 63, but there is no option of the answer

Step-by-step explanation:

Please give me brainliest ;)

4 0

3 years ago

Read 2 more answers

at a store, the probability that a customer buys socks is 0.15. the probability that a customer buys socks given that the custom

leva [86]

Answer: Option B

Step-by-step explanation:

First we assign a name to the events:

Event S: a customer buys socks

Event H: a customer buys shoes.

We know that :

$P (S) = 0.15$

We also know that the probability of S given that H occurs is:

$P(S|H) =\frac{P(S\ and\ H)}{P(H)}=0.20$

If two events S and H are independent then:

$P (S) * P (H) = P (S\ and\ H)$

This mean that if two events S and H are independent then:

$P(S|H) =\frac{P(S)*P(H))}{P(H)}$

$P(S|H) =P(S)$

We know that:

$P(S|H) =0.20$ and $P (S) = 0.15$

$0.20\neq 0.15$

This means that S and H events are dependent.

The answer is the option B

3 0

3 years ago

The life of a red bulb used in a traffic signal can be modeled using an exponential distribution with an average life of 24 mont

BartSMP [9]

Answer:

See steps below

Step-by-step explanation:

Let X be the random variable that measures the lifespan of a bulb.

If the random variable X is exponentially distributed and X has an average value of 24 month, then its probability density function is

$\bf f(x)=\frac{1}{24}e^{-x/24}\;(x\geq 0)$

and its cumulative distribution function (CDF) is

$\bf P(X\leq t)=\int_{0}^{t} f(x)dx=1-e^{-t/24}$

• What is probability that the red bulb will need to be replaced at the first inspection?

The probability that the bulb fails the first year is

$\bf P(X\leq 12)=1-e^{-12/24}=1-e^{-0.5}=0.39347$

• If the bulb is in good condition at the end of 18 months, what is the probability that the bulb will be in good condition at the end of 24 months?

Let A and B be the events,

A = “The bulb will last at least 24 months”

B = “The bulb will last at least 18 months”

We want to find P(A | B).

By definition P(A | B) = P(A∩B)P(B)

but B⊂A, so A∩B = B and

$\bf P(A | B) = P(B)P(B) = (P(B))^2$

We have

$\bf P(B)=P(X>18)=1-P(X\leq 18)=1-(1-e^{-18/24})=e^{-3/4}=0.47237$

hence,

$\bf P(A | B)=(P(B))^2=(0.47237)^2=0.22313$

• If the signal has six red bulbs, what is the probability that at least one of them needs replacement at the first inspection? Assume distribution of lifetime of each bulb is independent

If the distribution of lifetime of each bulb is independent, then we have here a binomial distribution of six trials with probability of “success” (one bulb needs replacement at the first inspection) p = 0.39347

Now the probability that exactly k bulbs need replacement is

$\bf \binom{6}{k}(0.39347)^k(1-0.39347)^{6-k}$

<em>Probability that at least one of them needs replacement at the first inspection = 1- probability that none of them needs replacement at the first inspection. </em>

This means that,

<em>Probability that at least one of them needs replacement at the first inspection = </em>

$\bf 1-\binom{6}{0}(0.39347)^0(1-0.39347)^{6}=1-(0.60653)^6=0.95021$

5 0

3 years ago

Can you help me with this question please

vaieri [72.5K]

To get the answer, divide 55 by 610. the answer would be 11.09 i’m pretty sure

5 0

3 years ago

Simplifly the expression 2(x - 5)

defon

Answer:

2x - 10

Step-by-step explanation:

2(x - 5)

distribute

2*x - 2*5

2x - 2*5

2x - 10

7 0

3 years ago

1f (-2, y) lies on the graph of y = 3, then y =​

1f (-2, y) lies on the graph of y = 3, then y =