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3 years ago

ASAP NEED HELP!!!!

Mathematics

2 answers:

padilas [110]3 years ago

7 0

Answer:

the answer should be 76.8 ounces

Step-by-step explanation:

Gnoma [55]3 years ago

7 0

Answer:

the answer should be 76.8 ounces

Step-by-step explanation:

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use the distributive property to remove parentheses. Simplify your answer as much as possible. 1/2(4y+7)

Anna007 [38]

1/2(4y+7)

Use the distributive property. a(b+c)= ab+ac

1/2*4y= 2y

1/2*7= 3.5

2y+3.5 <---- simplified expression

I hope this helps!
~kaikers

7 0

3 years ago

Read 2 more answers

Find h(2/3) if h(x) = 6x 2 4 6 8

arlik [135]

H(x) = 6x

it gives you what x is so plug that in the equation to find it.

h(2/3) = 6(2/3)
h(2/3) = 6 × 2 ÷ 3
h(2/3) = 12 ÷ 3
h(2/3) = 4

so your answer is 4.

hope this helps, God bless!

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

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

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

(5, 3) and (6, r) m = -1

Rzqust [24]

Answer:

r = 2

Step-by-step explanation:

(2-3)/(6-5)

= -1/1

= -1

6 0

3 years ago

A courier service advertises that its average delivery time is less than six hours for local deliveries. A random sample of size

Diano4ka-milaya [45]

Answer, Step-by-step explanation:

According to the exercise, we evaluate the delivery time of a courier company and we will hypothesize the best case with a sample size of 10, which is:

Small sample T test for single mean

The hypothesis that we will develop will be the following:

null hypothesis = mu> = 6

hypothesis alternativa: <6

3 0

3 years ago