Can someone solve this please? I do give brainliest points

Pavel [41]

We want to find the median for the given density curve.

The value of the median is 1.

Let's see how to solve this.

First, for a regular set {x₁, ..., xₙ} we define the median as the middle value. The difference between a set and a density curve is that the density curve is continuous, so getting the exact middle value can be harder.

Here, we have a constant density curve that goes from -1 to 3.

Because it is constant, the median will just be equal to the mean, thus the median is the average between the two extreme values.

Remember that the average between two numbers a and b is given by:

(a + b)/2

So we get:

m = (3 + (-1))/2 = 1

So we can conclude that the value of the median is 1, so the correct option is the second one, counting from the top.

If you want to learn more, you can read:

brainly.com/question/15857649

5 0

2 years ago

Use the mid-point rule with n = 4 to approximate the area of the region bounded by y = x3 and y = x. (10 points)

USPshnik [31]

See the graph attached.

The midpoint rule states that you can calculate the area under a curve by using the formula:
$M_{n} = \frac{b - a}{2} [ f(\frac{x_{0} + x_{1} }{2}) + f(\frac{x_{1} + x_{2} }{2}) + ... + f(\frac{x_{n-1} + x_{n} }{2})]$

In your case:
a = 0
b = 1
n = 4
x₀ = 0
x₁ = 1/4
x₂ = 1/2
x₃ = 3/4
x₄ = 1

Therefore, you'll have:
$M_{4} = \frac{1 - 0}{4} [ f(\frac{0 + \frac{1}{4} }{2}) + f(\frac{ \frac{1}{4} + \frac{1}{2} }{2}) + f(\frac{\frac{1}{2} + \frac{3}{4} }{2}) + f(\frac{\frac{3}{4} + 1} {2})]$
$M_{4} = \frac{1}{4} [ f(\frac{1}{8}) + f(\frac{3}{8}) + f(\frac{5}{8}) + f(\frac{7}{8})]$

Now, to evaluate your f(x), you need to look at the graph and notice that:
f(x) = x - x³

Therefore:
$M_{4} = \frac{1}{4} [(\frac{1}{8} - (\frac{1}{8})^{3}) + (\frac{3}{8} - (\frac{3}{8})^{3}) + (\frac{5}{8} - (\frac{5}{8})^{3}) + (\frac{7}{8} - (\frac{7}{8})^{3})]$

$M_{4} = \frac{1}{4} [(\frac{1}{8} - \frac{1}{512}) + (\frac{3}{8} - \frac{27}{512}) + (\frac{5}{8} - \frac{125}{512}) + (\frac{7}{8} - \frac{343}{512})]$

M₄ = 1/4 · (2 - 478/512)
= 0.2666

Hence, the <span>area of the region bounded by y = x³ and y = x</span> is approximately 0.267 square units.

6 0

3 years ago

Which expression is equivalent to 1/4(12x + 16) + 3x + 7

Nitella [24]

Answer:

the answer is 6x+11

Step-by-step explanation:

1/4(12+16)+3x+7

1/4×4(3x+4)+3x+7

(3x+4)+3x+7

3x+4+3x+7

6x+11

6 0

2 years ago

An individual who has automobile insurance from a certain company is randomly selected. Let y be the number of moving violations

Hoochie [10]

Answer:

a) $E(Y)= \sum_{i=1}^n Y_i P(Y_i)$

And replacing we got:

$E(Y) = 0*0.45 +1*0.2 +2*0.3 +3*0.05= 0.95$

b) $E(80Y^2) =80[ 0^2*0.45 +1^2*0.2 +2^2*0.3 +3^2*0.05]= 148$

Step-by-step explanation:

Previous concepts

In statistics and probability analysis, the expected value "is calculated by multiplying each of the possible outcomes by the likelihood each outcome will occur and then summing all of those values".

The variance of a random variable Var(X) is the expected value of the squared deviation from the mean of X, E(X).

And the standard deviation of a random variable X is just the square root of the variance.

Solution to the problem

Part a

We have the following distribution function:

Y 0 1 2 3

P(Y) 0.45 0.2 0.3 0.05

And we can calculate the expected value with the following formula:

$E(Y)= \sum_{i=1}^n Y_i P(Y_i)$

And replacing we got:

$E(Y) = 0*0.45 +1*0.2 +2*0.3 +3*0.05= 0.95$

Part b

For this case the new expected value would be given by:

$E(80Y^2)= \sum_{i=1}^n 80Y^2_i P(Y_i)$

And replacing we got

$E(80Y^2) =80[ 0^2*0.45 +1^2*0.2 +2^2*0.3 +3^2*0.05]= 148$

5 0

3 years ago

I need help with this questioun

andrey2020 [161]

Answer:

D

Step-by-step explanation:

8 0

2 years ago