I ALWAYS DO MY BROS WORK sO can u help him?

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

Kazeer [188]

Answer:

(1) (c) 5.30 years.

(2) (b) 0.289.

(3) (b) 0.80.

(4) (d) 0.50.

(5) (a) 5.25 years.

Step-by-step explanation:

Let X = age of the children in kindergarten on the first day of school.

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

The probability density function function of X 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 X 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) 5.30 years.

(2)

The standard deviation of a Uniform random variable is:

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

Compute the standard deviation of X 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) 0.289.

(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) 0.80.

(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) 0.50.

(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 x 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) 5.25 years.

6 0

3 years ago

Given: x ∥ y and w is a transversal Prove: ∠3 ≅ ∠6 Parallel lines x and y are cut by transversal w. On line x where it intersect

SOVA2 [1]

Answer:

The missing reason in the proof is transitive property

Step-by-step explanation:

Statement Reason

1. x ∥ y w is a transversal 1. given

2. ∠2 ≅ ∠3 2. def. of vert. ∠s

3. ∠2 ≅ ∠6 3. def. of corr. ∠s

4. ∠3 ≅ ∠6 4. ??????????

From the statements 2 and 3

The previous proved statement to make use of the transitive property reason or proof

∴ 4. ∠3 ≅ ∠6 4. transitive property

Note: the transitive property states that: If a = b and b = c, then a = c.

6 0

3 years ago

Read 2 more answers

At a book store there are 25 books on the clearance section. 20 of the books are young adult novels and the rest of the books ar

irga5000 [103]

The statement that is not true is A

6 0

3 years ago

Read 2 more answers

I'll give brainliest this is due soon so yeahmPLSSSSSSSSS

lora16 [44]

Answer:

He pays taxes on the amount after the discount. So 10% of $40.50 versus $45.

Step-by-step explanation:

10% on $40.50 = $4.05

10% on $45.00 = $4.50

4 0

3 years ago

Read 2 more answers

please help me with these problems.... please answer all the questions with the method I already started with... due at 11:59pm

dolphi86 [110]

Step-by-step explanation:

Howdy, did you do a, b, and c by yourself? Because there are some slight issues, so I'll do my best to explain.

a. 3x+5x

We know that x=2, so we can substitute 2 in for x right now, or we can combine like terms.

3x and 5x both have the variable x so we can add them easily to get 8x. If you substitute in 2, you're just multiplying 8×2, which is 16.

b. 5z+4z+z

Again, since

we know that z=-3, we can substitute in now, or we can combine like terms first. 5z, 4z, and z all have the variable x, we addition is easy. 5+4+1=10, so you'll have 10z. Substitute in -3 for z, and you're just multiplying 10×-3=-30

c. 3x+6+5x

This is trickier since 6 doesn't have the variable x. So to combine like terms means to only add 3x and 5x. So the equation will turn out looming like 8x+6. now substitute in 2 for x and you get

8(2)+6

16+6

22

d. 8y+8-4y

Combine like terms

4y+8

substitute in 5 for y

4(5)+8

20+8

28

e. 5z-4z+z

combine like terms

z+z

2z

substitute -3 for z

2(-3)

-6

f. 3x+6+5x-2

combine like terms

(so that's 3x+5x and 6-2)

8x+4

substitute in 2 for x

8(2)+4

16+4

20

g. 8y+8-4y-3

combine like terms

8y-4y+8-3

4y+5

substitute in 5 for y

4(5)+5

20+5

25

h. 5z-4z+z-3z

combine like terms

-z

substitute in -3 for z

-(-3)

double negatives make a positive

3

6 0

3 years ago