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2 years ago

1. Socks are on sale at 10% discount. What is the sale price of a pair of socks originally

Mathematics

1 answer:

Alona [7]2 years ago

4 0

Answer:

10/100 = x/125

1250 = 100x

1250/100 = 100x/100

12.5 = x

125 - 12.5 = 112.5

the sale price is $112.5

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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.

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

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

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.

This means that,

Probability that at least one of them needs replacement at the first inspection =

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

5 0

3 years ago

Simplify 11(3t+2)-4(2t-1

Ksenya-84 [330]

The answer to the question is 25t+26

6 0

2 years ago

Read 2 more answers

This is the graph of y=x^2+2x-2 what is the range of this function

Sedaia [141]

The range is [-3, ∞)

3 0

3 years ago

X - y = 2 4x – 3y = 11

swat32

Answer:

For x - y = 2

You need to find x or y in one of the equations and then substitute that into the other.

So we have;

x-y=2

4x-3y=11

We will take the first equation and find x;

x-y=2

add y to both sides;

x-y+y=2+y

x=2+y

Now we take that answer and substitute it forx in the other equation;

4(2+y)-3y=11

8+4y-3y=11

8+y=11

y=3

Now we have what y equals, so we use it in the first equation to find x;

x-3=2

x=5

So we have;

x=5; y=3

Hope you understand!

And for 4x – 3y = 11

Multiply the first equation by 2 and the second by 3 so that there are the same number of y's in each:

8x - 6y = 22 ...(3)

30x + 6y = -3 ...(4)

Now add (3) and (4) term by term:

38x + 0 = 19

38x = 19

or x = 1/2

Put this back into equation (1)

4*(1/2) - 3y = 11

2 - 3y = 11

Subtract 2 from both sides:

-3y = 9

Divide both sides by -3

y = -3

6 0

3 years ago

Read 2 more answers

Mary has used 160 tiles to cover 40% of the back splash in her kitchen how many tiles will it take to cover the entire back spla

dangina [55]

Answer:

400 tiles :)

Step-by-step explanation:

2/5 = 160/x

1/5 = 80/x

80 *5 = 400

6 0

3 years ago