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3 years ago

7

Find the greatest common factor of 84 and 36, showing how you got your answer. Then, using complete sentences, explain what your

answer means.

1 answer:

9966 [12]3 years ago

6 0

Answer:

Step-by-step explanation:

25 because im the sussest in the land

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If a worker can do his job in 5 hours, and a student can do the same job in 8 hours, what part of the work is left to finish aft

Aleks [24]

Answer:

Part of work left to finish after two work hours is 7/20

Step-by-step explanation:

Firstly, we need to calculate their joint work rate

That will be;

1/(1/5 + 1/8) = 1/(13/40) = 40/13

This means that they will complete the task in 40/13 hours

1 whole part takes 40/13

x part will take 2 hours

x * 40/13 = 2

40x = 26

x = 26/40

So the part that will be completed in two hours is 26/40

This means that the part left to complete will be:

1 - 26/40 = 14/40 = 7/20

8 0

2 years ago

Read 2 more answers

Lashonda sells beaded necklaces. Each large necklace sells for $6.60 and each small necklace sells for $4.90 . How much will she

xxMikexx [17]

46.2 dollars for the large necklaces and 14 dollars and seventy cents for the small necklaces therefore she will earn $60.9

5 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

Four friends participated in a free throw contest. Sean made twice as many

Wittaler [7]

Answer;

Sean got 24 points

Katie got 31 points

Kevin got 8 points

Emily got 12 points

Sorry I cannot give proper explanation because of typo issues.

Hope that can help.

4 0

2 years ago

22. Write the following as fractions: <br><br>i) Seven hundredths<br> ii) Six ninths

Brilliant_brown [7]

Here are both in fraction form.

i) 7/100
ii) 6/9

3 0

2 years ago