If g(x) = x+1/x-2 and h(x) = 4-x, what is the value of (g•h)(-3)?

If you when your mom went to the park and there were two dogs one dog went to the owner how many dogs are at the dog park

melomori [17]

Answer: 2 Dogs

Step-by-step explanation: The other dog never left the park

3 0

2 years ago

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Given rectangle QUAD, If DU = 5x - 4 and QP = 2x + 7, then find the length of DU.

ira [324]

Answer:

DU = 86

Step-by-step explanation:

Properties of a rectangle,

- Opposite sides are equal in measure.

- Diagonals of a rectangle are equal.

- Diagonals bisect each other.

In the given rectangle QUAD,

Diagonals AQ and DU bisect each other at P.

AQ = DU

2(QP) = DU [Since, AQ = 2(PQ)]

2(2x + 7) = 5x - 4

4x + 14 = 5x - 4

5x - 4x = 14 + 4

x = 18

Since, DU = 5x - 4

DU = 5(18) - 4

= 90 - 4

= 86

Therefore, measure of DU = 86 units

3 0

2 years ago

42-7=35 is an addend or equation

Virty [35]

42-7=35 is an equation

4 0

2 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

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

<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

2 years ago

anna owes home antiques Whose value is 160000 today assuming normal growth. however anna believes the value will grow at 16% per

Ray Of Light [21]

The formula is
A=p (1+r)^t
A future value?
P present value 160000
R interest rate 0.16
T time 3years
A=160,000×(1+0.16)^(3)
A=249,743.36

Use that future value to find the present value at a rate 8% compounded annually
To find p (present value) solve the formula for p
P=A÷ (1+r)^t
Where r is 0.08
P=249,743.36÷(1+0.08)^(3)
p=198,254.33

3 0

2 years ago