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3 years ago

Please help fast. 10 points.

Mathematics

2 answers:

Rudiy273 years ago

5 0

Answer:

commutative property

Step-by-step explanation:

you're welcome

Andrej [43]3 years ago

4 0

Answer: D

Step-by-step explanation:

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I reallly neeeeeeeeeeed helpppppppppppp

sdas [7]

Answer:

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

If you bought a tv for $350 and you bought it at a 10% discount off what was the original price

blagie [28]

(not sure if I'm right) Using this Formula Part/Whole = %/100

$350/Original Price = 10%/100
cross multiply
10w = 35,000
divide by 10 on both sides
W = $3,500
(seems real high so I'm not sure)

6 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

John wants to predict the frequency of rolling a 4 or 5 using a 6-sided dice with numbers one through six. Which of the followin

weeeeeb [17]

Answer: the answer is 120 or b

Step-by-step explanation: the answer is b

4 0

3 years ago

1f 2 sweaters cost $27.34, how much would 7 sweaters cost? The cost for 7 sweaters is s .

Mademuasel [1]

Answer:

95.69

Step-by-step explanation:

two sweaters costs 27.34 then one would cost 13.67

7×13.67 = 95.67

3 0

3 years ago