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3 years ago

Find the equation of the locus of a point, the sum of whose nisu

Mathematics

1 answer:

Ivenika [448]3 years ago

5 0

Answer: B or C sorry if its wrong and sorry if i got u wrong

Step-by-step explanation:

have a nice day

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Macy makes 15 of 20 free throws at basketball practice. What is the experimental probability that she will miss her next free th

My name is Ann [436]

Step-by-step explanation:

she might lose 1/4 of the next free throws

7 0

3 years ago

A number picked at random from the numbers 1 through 15 is prime. This "question" is on my study guide for Unit 6 Probability an

Sphinxa [80]

Solution:

we are given that

A number picked at random from the numbers 1 through 15 is prime.

As we know the number between 1 and 15 which are primes are listed below

2,3,5,7,11,13.

Here total number of favourable outcomes are 6.

The numbers 1 through 15 are as below:

1,2,3,4,5,6,7,8,9,10,11,12,13,14,15.

Here total number of possible outcoes are 15.

Hence the probability that if a number picked at random from the numbers 1 through 15 is prime $=\frac{6}{15}=\frac{2}{5}=0.40$

Hence the probability that if a number picked at random from the numbers 1 through 15 is prime is 0.4

6 0

3 years ago

Read 2 more answers

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>

$\bf 1-\binom{6}{0}(0.39347)^0(1-0.39347)^{6}=1-(0.60653)^6=0.95021$