Re-write the quadratic function below in Standard Form y = -9(x

What Is 8×6+4-4+8-8+6-6+8-8

lukranit [14]

The answer for this is 48.

4 0

3 years ago

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Volume of a pyramid height of 24ft base of 16ft

ratelena [41]

Answer:

sorry i guessed lol 384

Step-by-step explanation:

3 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

Can anyone solve 9, 10, 11, 12, and 13 I'm stuck on those problems, I already solved 14, 15, and 16 on paper.

ddd [48]

Answer:

9. x>-4 or x≥1

10. a<2 or a≥-5

11. v≤7 or v≥-4

12. k≥5 or k<8

13. n>6.8 and n≤9

Step-by-step explanation:

9. -2x-7>1

-2x>8

x>-4

x-2≥-1

x≥1

10.a/-2 <-1

a<2

-4a+3≥23

-4a≥20

a≥-5

11. 6v+38≤-4

6v≤-42

v≤7

2(v+3)≥-2

2v+6≥-2

2v≥-8

v≥-4

12. 4(1-k)≥-16

4-4k≥-16

-4k≥-20

k≥5

7-6k<-41

-6k<-48

k<8

13. 10n-9>-59

10n>-68

n>6.8

n-6≤3

n≤9

4 0

2 years ago

Find all values of x in the interval [0, 2π] that satisfy the equation. 6sin²(x) = 3

34kurt

Answer:

The solutions are π/4, 3π/4,5π/4,7π/4

Step-by-step explanation:

The given equation is

6sin²(x) = 3

Divide by 6 to get:

${ \sin}^{2} (x) = \frac{1}{2}$

This implies that;

$\sin(x) = \pm \frac{ \sqrt{2} }{2}$

If

$\sin(x) = \frac{ \sqrt{2} }{2}$

$x = \frac{\pi}{4}$

in the first quadrant

$x = \frac{3\pi}{4}$

in the second quadrant.

If

$\sin(x) = - \frac{ \sqrt{2} }{2}$

$x = \frac{5\pi}{4}$

in the third quadrant

$x = \frac{7\pi}{4}$

5 0

3 years ago

Re-write the quadratic function below in Standard Form y = -9(x - 1)(x + 2)