Given: △ABC, m∠A=60°, m∠C=45°, AB=9 Find: Perimeter of △ABC, Area of △ABC

jekas [21]

Given: In the given figure, there are two equilateral triangles having side 50 yards each and two sectors of radius (r) = 50 yards each with the sector angle θ = 120°

To Find: The length of the park's boundary to the nearest yard.

Calculation:

The length of the park's boundary (P) = 2× side of equilateral triangle + 2 × length of the arc

or, (P) = 2× 50 yards + 2× (2πr) ( θ ÷360°)

or, (P) = 2× 50 yards + 2× (2×3.14× 50 yards) ( 120° ÷360°)

or, (P) = 100 yards + 2× (2×3.14× 50 yards) ( 120° ÷360°)

or, (P) = 100 yards + 209.33 yards

or, (P) = 309.33 yards ≈309 yards

Hence, the option D:309 yards is the correct option.

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

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

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

Please help What is the value of x? A:12 B:15 C: 6 D: 30

pickupchik [31]

Answer: A, 12

Step-by-step explanation:

The line in the middle is a midsegment, making the two triangles similar by a ratio of two. So, dividing 60 by five will give you twelve

4 0

3 years ago

45% of $360 is what as a quantity

Bess [88]

I'm assuming you want to know what 45% of 360 is.....

so you multiply 360 by 0.45 (since 45 is a percent, and to revert it back, you must move the decimal 2 times to the left) and you have your answer...

5 0

3 years ago

You and three friends own a paint shop. The shop's profit was $536 in the first month. You always divide the profits equally. In

d1i1m1o1n [39]

C. $360 $536+$360=$896 $896/4= $360

3 0

3 years ago