All categories

3 years ago

PLEASE HELP ME

Mathematics

1 answer:

Andru [333]3 years ago

7 0

8 people have finished the race in 30 minutes

You might be interested in

Suppose the proportion X of surface area in a randomly selected quadrat that is covered by a certain plant has a standard beta d

dem82 [27]

Answer:

(a) The value of E (X) is 4/7.

The value of V (X) is 3/98.

(b) The value of P (X ≤ 0.5) is 0.3438.

Step-by-step explanation:

The random variable X is defined as the proportion of surface area in a randomly selected quadrant that is covered by a certain plant.

The random variable X follows a standard beta distribution with parameters α = 4 and β = 3.

The probability density function of X is as follows:

$f(x) = \frac{x^{\alpha-1}(1-x)^{\beta-1}}{B(\alpha,\beta)} ; \hspace{.3in}0 \le x \le 1;\ \alpha, \beta > 0$

Here, B (α, β) is:

$B(\alpha,\beta)=\frac{(\alpha-1)!\cdot\ (\beta-1)!}{((\alpha+\beta)-1)!}$

$=\frac{(4-1)!\cdot\ (3-1)!}{((4+3)-1)!}\\\\=\frac{6\times 2}{720}\\\\=\frac{1}{60}$

So, the pdf of X is:

$f(x) = \frac{x^{4-1}(1-x)^{3-1}}{1/60}=60\cdot\ [x^{3}(1-x)^{2}];\ 0\leq x\leq 1$

(a)

Compute the value of E (X) as follows:

$E (X)=\frac{\alpha }{\alpha +\beta }$

$=\frac{4}{4+3}\\\\=\frac{4}{7}$

The value of E (X) is 4/7.

Compute the value of V (X) as follows:

$V (X)=\frac{\alpha\ \cdot\ \beta}{(\alpha+\beta)^{2}\ \cdot\ (\alpha+\beta+1)}$

$=\frac{4\cdot\ 3}{(4+3)^{2}\cdot\ (4+3+1)}\\\\=\frac{12}{49\times 8}\\\\=\frac{3}{98}$

The value of V (X) is 3/98.

(b)

Compute the value of P (X ≤ 0.5) as follows:

$P(X\leq 0.50) = \int\limits^{0.50}_{0}{60\cdot\ [x^{3}(1-x)^{2}]} \, dx$

$=60\int\limits^{0.50}_{0}{[x^{3}(1+x^{2}-2x)]} \, dx \\\\=60\int\limits^{0.50}_{0}{[x^{3}+x^{5}-2x^{4}]} \, dx \\\\=60\times [\dfrac{x^4}{4}+\dfrac{x^6}{6}-\dfrac{2x^5}{5}]\limits^{0.50}_{0}\\\\=60\times [\dfrac{x^4\left(10x^2-24x+15\right)}{60}]\limits^{0.50}_{0}\\\\=[x^4\left(10x^2-24x+15\right)]\limits^{0.50}_{0}\\\\=0.34375\\\\\approx 0.3438$

Thus, the value of P (X ≤ 0.5) is 0.3438.

8 0

3 years ago

Point X has coordinates (-4, -5) and point y has coordinates (--1, --5 ) which expression models the distance between points X a

sdas [7]

Answer: The last expression shows the result of 3 so it models the distance between x and y.

Step-by-step explanation:

First you will have to find the distance between the two points by square the x coordinates and squaring the y coordinates to find their sum .

-4 - (-1) = -3

-5 - ( -5) = 0

-3^2 + 0^2 = d^2 where d is the distance

9 = d^2

d= 3

The distance is 3 units

Now evaluate which expression have a solution as 3.

The last expression is the answer

the absolute value for -4 is 4 , and the absolute value for -1 is 1 and 4 minus 1 is 3.

4 0

3 years ago

Identify the angles of elevation and angles of depression

Nadya [2.5K]

Answer:

1 is depression

2 is elevation

3 is depression

4 is elevation

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

3 different problems.

igor_vitrenko [27]

Answer:

see below

Step-by-step explanation:

b - 4 = 3

Add 4 to each side

b - 4+4 = 3+4

b = 7

-32 = 8w

Divide by 8

-32/8 = 8w/8

-4 = w

n/7 = -4

Multiply by 7 to each side

n/7 *7 = -4*7

n = -28

8 0

3 years ago

Read 2 more answers

A quadratic function and an exponential function are graphed below. How do the decay rates of the functions compareover the inte

Kobotan [32]

To check the decay rate, we need to check the variation in y-axis.

Since our interval is

$-2We need to evaluate both function at those limits.At x = -2, we have a value of 4 for both of them, at x = 0 we have 1 for the exponential function and 0 to the quadratic function. Let's call the exponential f(x), and the quadratic g(x).[tex]\begin{gathered} f(-2)=g(-2)=4 \\ f(0)=1 \\ g(0)=0 \end{gathered}$

To compare the decay rates we need to check the variation on the y-axis of both functions.

$\begin{gathered} \Delta y_1=f(-2)-f(0)=4-1=3 \\ \Delta y_2=g(-2)-g(0)=4-0=4 \end{gathered}$

Now, we calculate their ratio to find how they compare:

$\frac{\Delta y_1}{\Delta y_2}=\frac{3}{4}$

This tell us that the exponential function decays at three-fourths the rate of the quadratic function.

And this is the fourth option.

4 0

1 year ago