Dan says he ran twice as far as Karim. Give three possibilities for the distances each could have run.

Find the missing angle. 100 60 68 45 Will mark brain list !

Temka [501]

I think 45 because if you look at the top of the other 45 the graphs look the same lengths except this one stretches out so 45°

5 0

2 years ago

Tell whether the angles are adjacent or vertical. Then find the value of x.

maxonik [38]

Answer:

x = 63°

Step-by-step explanation:

~ They are adjacent angles because they have a common (one) vertex.

~ To find the value of x we must make an equation.

Angles on a straight line add up to 180°.

117 + x = 180

x = 180 - 117

x = 63°

8 0

3 years ago

A school wants to attract more students to its media center. The media specialist has been asked to conduct a survey to determin

aliina [53]

I would say C or B but I may be wrong

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

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

a survey asked 72 randomly chosen students if they were going to attend the school play. twelve said yes. if there are 210 ticke

Ugo [173]

We can solve this by setting up a proportion:

12 seeing the play/ 72 students = 210 seeing the play/ x students

We can cross multiply to solve for x:

(12)x=(72)(210)
12x=15,120
x=1,260 students who attend the school

4 0

3 years ago