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zhuklara [117]
4 years ago
15

77 multiply 52= . . . . . . . .. .. . . . . . . . . . ..

Mathematics
1 answer:
iogann1982 [59]4 years ago
6 0
1.48 is the answer to ur question 
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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
Kazeer [188]

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>.

6 0
3 years ago
(Will mark as brainly) please show ur work on how u got ur answer
elixir [45]

Answer:

1.=6  2.=7.6 3.= 2.2 6.= 3.50

Step-by-step explanation:

1. 78-27 =51 51 divided by 8.50=6

2.150-27=123 123 divided by 16=7.6

3. 14-65=51 51 divided by 23=2.2

6. 10 divided by 3= 3.5

7 0
3 years ago
What is information given in numbers called? A) facts B)data C) figures D)estimates
yKpoI14uk [10]

Answer:

B) Data

Step-by-step explanation:

Data has to do with numbers.

3 0
4 years ago
Read 2 more answers
A football is kicked into the air from an initial height of 4 feet. The height, in feet, of the football above the ground is giv
exis [7]
25 = -16t^2 + 50t + 4
16t^2 - 50t + 21 = 0
16t^2 - 8t - 42t + 21 = 0
8t(2t - 1) - 21(2t - 1) = 0
(8t - 21)(2t - 1) = 0
8t - 21 = 0 or 2t - 1 = 0
t = 21/8 or t = 1/2
t = 2.625 or t = 0.5
7 0
4 years ago
Read 2 more answers
3 and a Half cubes of sand transported at once. How many times do vehicle travel to transport 28 sand cubes. ​
Tema [17]

Answer:

8 times

Step-by-step explanation:

28/3.5 = 8

8 0
3 years ago
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