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Taya2010 [7]
2 years ago
10

The longer base of a trapezoid is 8 ft. The longer base of a similar trapezoid is 13 ft. The area of the smaller trapezoid is 24

0 ft2. What is the area of the larger trapezoid?
Mathematics
1 answer:
Licemer1 [7]2 years ago
3 0

Answer:

390 ft²

Step-by-step explanation:

The longer base of a trapezoid is 8 ft. The longer base of a similar trapezoid is 13 ft. The area of the smaller trapezoid is 240 ft² What is the area of the larger trapezoid?

We solve the above question using proportion

(Longer base/Area of trapezoid) smaller trapezoid = (Longer base/Area of trapezoid) bigger trapezoid

Let the the Area of the bigger trapezoid = x

Hence,

= 8ft/240ft = 13ft/x ft

Cross Multiply

8ft × x = 240ft × 13ft

x = 240ft² × 12 ft/8 ft

x = 390 ft²

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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
If x = -2, then x2-7x+10 equals<br> 00<br> O 20<br> 028
sleet_krkn [62]
20, you just need to substitute the x value in the equation
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7 0
3 years ago
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maksim [4K]
Answer:C


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8 0
2 years ago
Question 2.what is the probability that between 80 and 115 complaints are received in one week? use a n(110,20) model to approxi
AURORKA [14]

Calculate standard  Z values for the given data:-

z = (80 - 110) / 20 = - 3/2 = -1.5

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6 0
3 years ago
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Arlecino [84]

Answer:

C

Step-by-step explanation:

2x + 1 = 85 (Alternate Interior Angles Theorem)

2x = 84

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3y + 5 + (85) = 180

3y + 5 = 95

3y = 90

<em>y = 30</em>

7 0
2 years ago
Read 2 more answers
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