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3 years ago

What is 61.12÷3.2 uses to expressions.

Mathematics

1 answer:

Sergeu [11.5K]3 years ago

5 0

Answer:

answer is 19.1

Step-by-step explanation:

answer is 19.1

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Please answer correctly !!!!!!!!!!!!!!!!!! Will mark brainliest !!!!!!!!!!!!!!!!!

Olin [163]

U is a point in the line segment TV

TU + UV = TV

TU = 12

UV = 3

Hope this will help...

4 0

3 years ago

The isosceles triangle has a perimeter of 7.5 m. Which equation can be used to find the value of x if the shortest side, y, meas

kicyunya [14]

An isosceles triangle has 2 sides of the same length and one side that's not the same length.

So it says the shortest side, y, is 2.1m, so that means it's the side that is not the same length as the other two sides.

But the other two sides will be the same length. Let's make those 2 sides x.

The equation would be
x + x + 2.1 = 7.5

And with that you can find x.

3 0

3 years ago

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NARA [144]

Answer:

sorry can't help hope someone gets it

6 0

3 years ago

6 feet 2 inch statue I want it to be taller by a ratio of 2:1 how tall will it be

Sholpan [36]

Answer:

Step-by-step explanation:

2* (6' 2") = 12' 4" will be the height of the taller statue.

7 0

3 years ago

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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

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

<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