Explain why the terms 4x and 4x^2 are not like like terms

A car is traveling up a steep ramp to a parking garage.the ramp is 445 feet long and rises a vertical distance of 80 feet. What

Alecsey [184]

With the information provided in the problem, we can create a right triangle with the ramp as its hypotenuse, and the vertical rise as its opposite side from its angle of elevation.
Let $E$ be the angle of elevation from the car to the end of the ramp. We now know that the hypotenuse of our triangle measures 445 feet, and the opposite side measures 80 feet, so we need a trig function that relates our angle of elevation with the hypotenuse and the opposite side. That function is sine:
$sine(E)= \frac{80}{445}$
$E=arcsine( \frac{80}{445} )$
$E=10.36$

We can conclude that the angle of elevation from the car to the end of the ramp is 10.36°.

3 0

3 years ago

Read 2 more answers

What is x+y=8 and 5x+3y=34 graphed on the same graph

expeople1 [14]

To help you graph this, get y alone in both equations. If you don't have a graphing calculator, use the Desmos graphing calculator online.

So, you will end up graphing this:
y=8-x
y=(34/3)-(5/3)x

They will meet at (5,3)

8 0

3 years ago

Can 8/45 be simplified as a fraction

max2010maxim [7]

Answer:

Step-by-step explanation:

isn't it already a fraction or is it a division

if it's a division then yes

the answer is 5 and 2/8

8 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

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

<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>

$\bf 1-\binom{6}{0}(0.39347)^0(1-0.39347)^{6}=1-(0.60653)^6=0.95021$