All categories

3 years ago

I need help with this question.

Mathematics

2 answers:

Bumek [7]3 years ago

5 0

C I agree with that person

mart [117]3 years ago

4 0

It’s C give me like and brainleist please

You might be interested in

Directions: Apply me u you answer should be w 1) 12. 12² =

makkiz [27]

12×12 squared is = 12•144/=

1728

If you need a better explanation just ask in the comments.

Happy to help! Please mark as BRAINLIEST! Thanks

7 0

4 years ago

2,10,18,26 find the 61st term

saw5 [17]

Answer:

482

Step-by-step explanation:

We can see that the numbers shown resemble an arithmetic sequence because they have a common difference. The formula for the nth term of an arithmetic sequence is:

$a_{n} =a_{1} +(n-1)d$

Where $a_{1}$ is the first term, $a_{n}$ is the nth term, and $d$ is the common difference. To find the 61st term, all we need is the first term and the common difference. By looking at what given, we can say the first term is 2. Now, to find the common difference, we find the difference of a term from the term before it. In this case we can do $10-2$ , which is $8$ , or the common difference. Since we have everything we need, it can be plugged into the equation:

$a_{61} =2 + (61 - 1)*8\\a_{61} = 2 + 60*8\\a_{61} = 2 + 480\\a_{61} = 482$

So, the 61st term is 482.

4 0

3 years ago

Please resolve this is for today

Fynjy0 [20]

Answer:

A Square plus B square equals

Step-by-step explanation:

5 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

Factor the expession using the gcf. 20p-60=

kotegsom [21]

That'd be 20(p-3). Check by multiplying out 20(p-3)

5 0

3 years ago

I need help with this question.​

I need help with this question.