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3 years ago

Which faces have an area of 28in squared

Mathematics

2 answers:

madreJ [45]3 years ago

7 0

The lowest rectangle is 28in because you just multiply the 4 and the 7

kvasek [131]3 years ago

3 0

The very last square the one with 4in

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26 out of 40 whats the percent

Mademuasel [1]

26 out of 40 is the same as 26 divided by 40 and it is 0.65. To make a decimal into a percent, you multiply it by 100 and the the percent sign: 0.65*100= 65%
Therefore, 26 out of 40 is 65%

6 0

3 years ago

Read 2 more answers

Dan invests £13000 into his bank account. He receives 2.3% per year simple interest. How much will Dan have after 4 years? Give

frozen [14]

Answer:

=14237.90

Step-by-step explanation:

13000*1.023^4=14237.89832

=14237.90

8 0

3 years ago

ILL GIVE BRAINLYIST AND ALL MY POINTS IF THIS GETS ANSWERD JUST HURRY!!!!

Step2247 [10]

Answer:

0.261

Step-by-step explanation:

gimme brainliest

4 0

2 years ago

DUE TONIGHT!!! 20 POINTS Graph 3x - y = .2 A) B) C) D)

alisha [4.7K]

I believe the answer is A

6 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

devlian [24]

Answer:

a) 0.39347

b) 0.22313

c) 0.95021

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 atleast 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

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

We also have:

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 =

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

4 0

3 years ago