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3 years ago

At the store you can get 15 cans of peaches for 3 dollars find the unit rate for 1

Mathematics

1 answer:

labwork [276]3 years ago

3 0

You can get 5 cans per dollar

And is the same as saying 15 cans of peaches divided by 3 dollars

-- Actually i'm pretty confused about the question so i'll come back to this later

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What’s nine plus ten

natka813 [3]

Answer:

Step-by-step explanation:

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3 years ago

PLEASE HELP AND THANK YOU!! I WILL ALSO MARK BRAINLIEST IF THE ANSWER IS CORRECT!!

Sergeeva-Olga [200]

Yes you are right! There is much more balance in the left than in the right!

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3 years ago

Read 2 more answers

The sum to infinity of a geometric sequence is twice the first term. What is the common ratio?

Dovator [93]

Answer:

The constant factor between consecutive terms of a geometric sequence is called the common ratio.

Example:

To find the common ratio , find the ratio between a term and the term preceding it. r=42=2. 2 is the common ratio.

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3 years ago

on monday, steve read 9 pages of his new book. to finish the first chapter on tuesday, he needs to read double the number of pag

Greeley [361]

18 pages because nine needs to be doubled

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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

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

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