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2 years ago

Plz help. I will mark Brainlyist

Mathematics

1 answer:

Katarina [22]2 years ago

3 0

Answer:

the answer is 2hsox864

Step-by-step explanation:

so I'd you with all the way I wasn't really going on the road to be there today because I have to be with my mom said to me that was the first one I've been to know you can do that is weird but you have a lot to say that I have to say that you have to do it this is what I sent you and you can be a witch and you will be able and I dont you are so

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Answer in factored form 13+14m+m^2

natta225 [31]

Solution

m = {-13, -1}

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3 years ago

Read 2 more answers

An actor is 6ft tall.On a billboard for a new movie,the actors pictures is enlarged so that his height is 16.8 ft.what is the sc

Darya [45]

Answer:

Scale Factor = 2.8

Step-by-step explanation:

The scale factor is how many times LARGER is the enlarged pic.

We can divide the enlarged pic (16.8) by original (6) to get the scale factor:

Scale Factor = $\frac{16.8}{6}=2.8$

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3 years ago

Anita ran 3 miles in 33 minutes. Write and solve a multiplication equation to find out how long it took Anita to run each mile i

zhuklara [117]

It took 11 minutes because 3 times 11=33

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3 years ago

The median of a probability distribution can be defined as the number m such that Upper P (Upper X less than or equals m )equals

Alexxandr [17]

Answer:

tex]M=\beta ln(2)[/tex]

Step-by-step explanation:

Previous concepts

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate).

Solution to the problem

For this case we can use the following Theorem:

"If X is a continuos random variable of the exponential distribution with parameter $\beta$ for some $\beta \in R >0$ "

Then the median of X is $\beta ln (2)$

Proof

Let M the median for the random variable X.

From the definition for the exponential distribution we know the denisty function of X is given by:

$f_X (x) = \frac{1}{\beta} e^{-\frac{x}{\beta}}$

Since we need the median we can put this equation:

$P(X$

If we evaluate the integral we got this:

$\frac{1}{\beta} \int_0^M e^{- \frac{x}{\beta}}dx =\frac{1}{\beta} [-\beta e^{-\frac{x}{\beta}}] \Big|_0^M$

And that's equal to:

$1/2 = 1 -e^{- \frac{M}{\beta}}$

And if we solve for M we got:

$1-e^{- \frac{M}{\beta}} = \frac{1}{2}$

$e^{- \frac{M}{\beta}}=\frac{1}{2}$

If we apply natural log on both sides we got:

$-\frac{M}{\beta}=ln(1/2)$

And then $M=\beta ln(2)$

5 0

3 years ago

4. A small high school holds its graduation ceremony in the gym. Because of seating constraints, students are limited to a maxim

Ad libitum [116K]

Answer:

(a) The mean and standard deviation of X is 2.6 and 1.2 respectively.

(b) The mean and standard deviation of T are 390 and 180 respectively.

(c) The distribution of T is N (390, 180²). The probability that all students’ requests can be accommodated is 0.7291.

Step-by-step explanation:

(a)

The random variable X is defined as the number of tickets requested by a randomly selected graduating student.

The probability distribution of the number of tickets wanted by the students for the graduation ceremony is as follows:

X P (X)

0 0.05

1 0.15

2 0.25

3 0.25

4 0.30

The formula to compute the mean is:

$\mu=\sum x\cdot P(X)$

Compute the mean number of tickets requested by a student as follows:

$\mu=\sum x\cdot P(X)\\=(0\times 0.05)+(1\times 0.15)+(2\times 0.25)+(3\times 0.25)+(4\times 0.30)\\=2.6$

The formula of standard deviation of the number of tickets requested by a student as follows:

$\sigma=\sqrt{E(X^{2})-\mu^{2}}$

Compute the standard deviation as follows:

$\sigma=\sqrt{E(X^{2})-\mu^{2}}\\=\sqrt{[(0^{2}\times 0.05)+(1^{2}\times 0.15)+(2^{2}\times 0.25)+(3^{2}\times 0.25)+(4^{2}\times 0.30)]-(2.6)^{2}}\\=\sqrt{1.44}\\=1.2$

Thus, the mean and standard deviation of X is 2.6 and 1.2 respectively.

(b)

The random variable T is defined as the total number of tickets requested by the 150 students graduating this year.

That is, T = 150 X

Compute the mean of T as follows:

$\mu=E(T)\\=E(150\cdot X)\\=150\times E(X)\\=150\times 2.6\\=390$

Compute the standard deviation of T as follows:

$\sigma=SD(T)\\=SD(150\cdot X)\\=\sqrt{V(150\cdot X)}\\=\sqrt{150^{2}}\times SD(X)\\=150\times 1.2\\=180$

Thus, the mean and standard deviation of T are 390 and 180 respectively.

(c)

The maximum number of seats at the gym is, 500.

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sum of values of X, i.e ∑X, will be approximately normally distributed.

Here T = total number of seats requested.

Then, the mean of the distribution of the sum of values of X is given by,

$\mu_{T}=n\times \mu_{X}=390$

And the standard deviation of the distribution of the sum of values of X is given by,

$\sigma_{T}=n\times \sigma_{X}=180$

So, the distribution of T is N (390, 180²).

Compute the probability that all students’ requests can be accommodated, i.e. less than 500 seats were requested as follows:

$P(T$

Thus, the probability that all students’ requests can be accommodated is 0.7291.

8 0

3 years ago