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melamori03 [73]
3 years ago
7

NEED ASAP plz. I will also mark brainiest! Don't just take points, I will report you.

Mathematics
1 answer:
PilotLPTM [1.2K]3 years ago
3 0

Part A:

You may choose the two lines connecting the origin and points A and B, and choose the portion of the space between them.

The line between the origin and A is

y = 3x

We want everything below this line (line included), so the first inequality is

y \leq 3x

The line between the origin and B is

y = \dfrac{1}{3}x

We want everything above this line (line included), so the second inequality is

y \geq \dfrac{1}{3}x

Create a system with these two inequalities and you'll have an area including only points A and B

Part B:

To verify the solutions, we can plug the coordinates of A and B in this system and check that we get something true: the coordinates of point A are (1,3), while the coordinates of point B are (3,1). The system becomes:

A:\begin{cases}3 \leq 3\cdot 1\\3 \geq \frac{1}{3}\cdot 1\end{cases},\quad B:\begin{cases}1 \leq 3\cdot 3\\1 \geq \frac{1}{3}\cdot 3\end{cases}

Which means

A:\begin{cases}3 \leq 3\\3 \geq \frac{1}{3}\end{cases},\quad B:\begin{cases}1 \leq 9\\1 \geq 1\end{cases}

And these are all true. So, the system is satisfied, which means that the points belong to the shaded area.

Part C

If you draw the line, you'll see that the only points that lay below the line are B and C. In fact, if we plug the coordinates we have

B:\ 1

And this are both true. You can check the coordinates of all other points, and see that they won't satisfy the inequality y<3x-6

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The age of the children in kindergarten on the first day of school is uniformly distributed between 4.8 and 5.8 years old. A fir
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Answer:

(1) (c) <u>5.30 years</u>.

(2) (b) <u>0.289</u>.

(3) (b) <u>0.80</u>.

(4) (d) <u>0.50</u>.

(5) (a) <u>5.25 years</u>.

Step-by-step explanation:

Let <em>X</em> = age of the children in kindergarten on the first day of school.

The random variable <em>X</em> follows a continuous Uniform distribution with parameters <em>a</em> = 4.8 years and <em>b</em> = 5.8 years.

The probability density function function of <em>X</em> is:

f_{X}(x)=\left \{ {{\frac{1}{b-a}} ;\ a

(1)

The expected value of a Uniform random variable is:

E(X)=\frac{1}{2}(a+b)

Compute the mean of <em>X</em> as follows:

E(X)=\frac{1}{2}(a+b)=\frac{1}{2}\times (4.8+5.8)=5.3

Thus, the  mean of the distribution is (c) <u>5.30 years</u>.

(2)

The standard deviation of a Uniform random variable is:

SD(X)=\sqrt{\frac{1}{12}(b-a)^{2}}

Compute the standard deviation of <em>X</em> as follows:

SD(X)=\sqrt{\frac{1}{12}(b-a)^{2}}=\sqrt{\frac{1}{12}\times (5.8-4.8)^{2}}=0.289

Thus, the standard deviation of the distribution is (b) <u>0.289</u>.

(3)

Compute the probability that a randomly selected child is older than 5 years old as follows:

P(X>5)=\int\limits^{5.8}_{5} {\frac{1}{5.8-4.8}}\, dx\\

                =\int\limits^{5.8}_{5} {1}\, dx\\=[x]^{5.8}_{5}\\=(5.8-5)\\=0.8

Thus, the probability that a randomly selected child is older than 5 years old is (b) <u>0.80</u>.

(4)

Compute the probability that a randomly selected child is between 5.2 years and 5.7 years old as follows:

P(5.2

                            =\int\limits^{5.7}_{5.2} {1}\, dx\\=[x]^{5.7}_{5.2}\\=(5.7-5.2)\\=0.5

Thus, the probability that a randomly selected child is between 5.2 years and 5.7 years old is (d) <u>0.50</u>.

(5)

It is provided that a randomly selected child is at the 45th percentile.

This implies that:

P (X < x) = 0.45

Compute the value of <em>x</em> as follows:

   P (X < x) = 0.45

\int\limits^{x}_{4.8} {\frac{1}{5.8-4.8}}\, dx=0.45

        \int\limits^{x}_{4.8} {1}\, dx=0.45

           [x]^{x}_{4.8}=0.45

       x-4.8=0.45\\

                x=0.45+4.8\\x=5.25

Thus, the age of the child at the 45th percentile is (a) <u>5.25 years</u>.

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Answer:

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Step-by-step explanation:

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