Help please?

Solve the inequality

const2013 [10]

X < -11/9 is the answer

4 0

2 years ago

What is the difference of 7 1/6−3 5/8? fractions. with common denominators

lord [1]

Answer: 3 13/24

Step-by-step explanation:

Turn 7 1/6 into an Improper Fraction:

43/6 - 3 5/8

Turn 3 5/8 into an Improper Fraction:

43/6 - 29/8

Subtract:

43/6 - 29/8= 85/24

Turn 85/24 into a Mixed Number (Final Step):

85/24 = 3 13/24

Answer:

3 13/24

Hope this Helps!

4 0

2 years ago

12+root of 35 divided by 2 + 12-2 root of 35 divided by 2

nadya68 [22]

So : $12+ \frac{\sqrt{35}}{2} + \frac{12-2 \sqrt{35} }{2}$ ?

First, you want to bring everything together on the top of the fraction.
12=24/2 so you can have $\frac{24+ \sqrt{35}+12-2 \sqrt{35} }{2}$

Add like terms -- 12 +24 = 36 $\sqrt{35}-2 \sqrt{35} = \sqrt{35}$

Now we are at $\frac{24+ \sqrt{35} }{2}$ Which is the lowest possible simplified version while remaining at exact value.

Feel free to check my math! I kinda did this off the top of my head.

8 0

2 years ago

Which values of P and Q result in an equation with no solutions? 83x+P=83x+Q

aleksandr82 [10.1K]

83x+P=83x+Q

subtract 83x from each side

P=Q

if P does not Q there are no solutions

the lines would be parallel and never intersect if Q does not equal P

6 0

2 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

2 years ago