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Ivenika [448]
3 years ago
15

Helpp do u know this it would be really helpful if u helped

Mathematics
1 answer:
Bogdan [553]3 years ago
6 0

Answer: y = 8*x + 5

Step-by-step explanation:

Here we have the points (1, 13), (2, 21), (3, 29), (4, 37)

You can see that the x-component always increases by one unit, while the y-component always increases by 8 units.

Then we can conclude that this is a linear relationship.

A linear relationship can be written as:

y = a*x + b

where a is the slope and b is the y-axis intercept.

For a line that passes through the points (x1, y1) and (x2, y2), the slope can be written as:

a = (y2 - y1)/(x2 - x1).

Here we can use the fact that our line passes through the points (1, 13) and (2, 21)

Then the slope of our line is:

a = (21 - 13)/(2 - 1) = 8

Then:

y = 8*x + b

To find the value of b, we can use again the fact that our line passes through the point (1, 13)

This means that when x = 1, the value of y is 13, if we replace that in the equation we get:

13 = 8*1 + b

13 = 8 + b

13 - 8 = b

5 = b

Then:

y = 8*x + 5

This means the amount charged for x hours, is:

y = 8*x + 5

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Jessica pays a membership fee of $40 so that's given to us <span>Jessica has a membership fee of $40 (given)
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Simplify (5 x + 1) - (-10 x + 6)
tekilochka [14]

Answer:15x+7

Step-by-step explanation:

First you would see that the -(-10x+6) would turn into 10x+6 meaning that the whole equation would turn into 5x+1+10x+6. Then add the x numbers together and the whole numbers separate. So now your answer should be 15x+7

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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.

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• 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)

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\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

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