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4 years ago

Helppppp mmmeeeee!!!! plzzzz

Mathematics

2 answers:

castortr0y [4]4 years ago

6 0

Answer:

96 in²

Step-by-step explanation:

SA=6(s²)

SA=6(4²)

SA=6(16)

SA=96

Softa [21]4 years ago

3 0

Answer:

A = 96 in.²

Step-by-step explanation:

Surface are of a cube: A = 6s²

s = 4 in. in this case.

A = 6(4)²

A = 6(16)

A = 96 in.²

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Which term describes the dotted line in the quadrilateral?

Tasya [4]

The correct answer is C diagonal

7 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>

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

5 0

3 years ago

Help and help on my recent question too please

sveticcg [70]

For the second one 1 mph equals 1.6 km so i’m pretty sure it would be all of them

5 0

3 years ago

Read 2 more answers

WILL MARK BRAINLIEST

atroni [7]

Answer:

7.405882353 years

Step-by-step explanation:

Simple interest is

A = P(1+rt)

Where A is the amount in the account

P is the principle invested

r is the rate and

t is the time

6593.75 = 5000( 1+ .0425*t)

Divide each side by 5000

6593.75/5000 = ( 1+ .0425*t)

1.31475 = ( 1+ .0425*t)

Subtract 1 from each side

.31475 = .0425t

Divide each side by.0425

.31475/.0425 = .0425t/.0425

7.405882353 = t

7.405882353 years

8 0

4 years ago

Read 2 more answers

Help help math math math

natka813 [3]

Answer:

Yes, this is a function

Step-by-step explanation:

You know this because it passes the vertical line test.

I hope this helps!

7 0

2 years ago