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3 years ago

Jim drank 2/5 of his water bottle and John drank 3/10 of his water bottle. How much did the both boys drink?

Mathematics

1 answer:

suter [353]3 years ago

4 0

The boys drink 7/10 of a water bottle

1. Convert 2/5 into 4/10
2. Add 4/10 and 3/10

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dimaraw [331]

32, i believe is the answer

5 0

4 years ago

Diameter of the circle whose radius is given a) 3.5cm

sdas [7]

To find the diameter of the radius you need to multiply 3.5 x 2 = 7

5 0

3 years ago

What is the area of this figure I need help please

coldgirl [10]

Answer:

Area = 126

Step-by-step explanation:

To figure out the area of the triangle, we must do 1/2 base times height. Which would be 1/2 of 6 times 6. Which is 18. Then to figure out the rectangle is just doing the width times the length and the same for the square. So 6 times 12 plus 6 times 6 equals 108. Then we add the triangles area and we come out with 108 plus 18 which equals 126.

3 0

4 years ago

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

What's the slope of (1,5) and (4, -2)

madreJ [45]

Step-by-step explanation:

Slope =y3 - y1 / x2 - x1

Put the values and the answer will come

3 0

3 years ago