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3 years ago

A herd of bison currently has 55 members. Based on the available resources,

Mathematics

2 answers:

mixer [17]3 years ago

5 0

Answer:

Written out the equation would look like f(x) = 55(1.06)^x

The graph would look something like the attached picture. (It intercepts with the y-axis at 55).

Marysya12 [62]3 years ago

5 0

Answer:

The graph should intersect at point (0,55). I attatched a screen shot to show what it should look like!

Step-by-step explanation:

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Find f^-1(x) for f(x) = 1/x^3

kow [346]

Answer: $f^{-1}(x) = \frac{\sqrt[3]{x^{2}}}{x}$

Step-by-step explanation:

$y = \frac{1}{x^{3}}$

Inverse is when you swap the x's and y's and then solve for "y":

$x = \frac{1}{y^{3}}$

$y^{3} = \frac{1}{x}$

$y = \frac{1}{\sqrt[3]{x}}$

$y = \frac{1}{\sqrt[3]{x}}*(\frac{\sqrt[3]{x}}{\sqrt[3]{x}})^{2}$

$y = \frac{\sqrt[3]{x^{2}}}{x}$

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Alex

That would be 9c+63 i’m pretty sure

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3 years ago

What is the relationship between the 6’s in 660,472

belka [17]

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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.

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3 years ago

What is a good word problem for 5(x+3)=170

Tatiana [17]

think of something and someone I can't really answer this question right now but if you want you can give me 1 star but I hope you get your question answered and have a WONDERFUL day

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