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HACTEHA [7]
3 years ago
7

What the two triangles have in common?!

Mathematics
1 answer:
pantera1 [17]3 years ago
4 0
I notice that both triangles have corresponding angles.
This means the triangles are similar.
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What is 16/20 simplified
Digiron [165]
<u>4/5</u>, you divide both by 4 and you get <u>4/5</u>
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The Statue of Liberty is approximately 300 ft. tall, and the right arm is approximately 42 ft. long. On a scale model that is 2
Archy [21]

Answer:

Weight of copper: 62,000 lbs. (31 tons). Weight of framework: 250,000 lbs. (125 tons). Weight of concrete foundation: 54,000,000 lbs. Thickness of copper sheeting: 3/32 of an inch, the thickness of two pennies placed together.  the answer in 125

Step-by-step explanation:

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Beth will you help Andy with his algebra?
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Answer:

A)

Step-by-step explanation:

mark me brainliest plzzzzzzz

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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
How does the graph of y=-3√2x-4 compare to the graph of y=-3√x-4?
kifflom [539]
<span>Assuming the graph is y=-3(√2x)-4 and y=-3√(x-4) the transformation would be:

</span><span>The graph is compressed horizontally by a factor of 2
x=1/2x'
</span>y=-3(√2x)-4 
y=-3(√x')-4 <span>

</span><span>moved left 4
x=x'-4
</span>y=-3(√x)-4 
y=-3(√x'-4)-4 
<span>
moved down 4
y=y'-4
</span>y=-3(√x-4)-4 
y'-4=-3(√x'-4)-4 
y'=-3(√x'-4)-4 +4
y'=-3(√x'-4)

Answer: C. <span>The graph is compressed horizontally by a factor of 2, moved left 4, and moved down 4.
</span>
6 0
3 years ago
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