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4 years ago

Expand the exponents, simplify, then write the rule.

Mathematics

1 answer:

worty [1.4K]4 years ago

5 0

3) if base are same power are added

4) power is multiplied

5) in division, if base are equal power is subtracted

6) we can write whole power if two bases have same power

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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) 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

–8 p + 4 u + 6, when p = 9 and u = 3 Ok at this point I'm double checking all of my answers lol

algol13

Answer:

-54

Step-by-step explanation:

-72+12+6

4 0

3 years ago

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mr hill has 27 students in his class and mr chang has 24 students in his class.both classes will be divided into equal sized tea

Svetlanka [38]

Answer:

mr hill: 9

mr chang: 12

Step-by-step explanation:

mr hill can have 3 teams of 9 and

mr chang can have 2 teams of 12

5 0

3 years ago

Read 2 more answers

Sean has two single accounts at Corona Bank. One account has a balance of $174,898.09 and the

klasskru [66]

$452898.08 is the answer

3 0

3 years ago

There are 100 seniors and 80 juniors in the Homecoming Parade. The students split up into groups of equal size, to fit on the pa

Masteriza [31]

Answer:

20 students should be on each float

Step-by-step explanation:

Number of seniors = 100

Number of juniors = 80

To find number of students that should be on each float, find highest common factor (H.C.F) of 100 and 80

Write prime factorisation of 100 and 80.

100 = $2^2$ × $5^2$

80 = $2^4$ × 5

So,

H.C.F(100, 80) = $2^2$ × 5 = 4 × 5 = 20

Therefore,

20 students should be on each float

3 0

3 years ago