1/5 divided by 3 is what

What is 60% of 52? (SHOW WORK)

klemol [59]

Answer:

31.2

Step-by-step explanation:

i googled it

5 0

2 years ago

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Total consumption of fruit juice in a particular country in 2006 was about 1.85 billion gallons. The population of that country

Eduardwww [97]

Given :

Population of country in 2006 , P = 300 million.

Consumption of fruit juice in 2006 by while population, C = 1.85 billion.

To Find :

The average number of gallons of fruit juice consumed per person in the country in 2006.

Solution :

We know, 1 billion = 1000 million.

C = 1.85 billion = 1850 million.

Now, average umber of gallons of fruit juice consumed per person in the country in 2006 is :

$c =\dfrac{C}{P}\\\\c=\dfrac{1850}{300}\ gallons/person\\\\c=6.17 \ gallons/person$

Hence, this is the required solution.

4 0

2 years ago

the two shorter sides of a triangle are the same length. the length of the longer side is 5 m longer than each of the shorter si

STatiana [176]

Let the length of the shorter sides be x, then
Perimeter = x + x + x + 5
29 = 3x + 5
3x = 29 - 5 = 24
x = 24/3 = 8

Therefore, the length of longest side is 8 + 5 = 13 m.

4 0

2 years ago

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A contractor is required by a county planning department to submit one, two, three, four, or five forms (depending on the nature

Westkost [7]

Answer:

(a) The value of k is $\frac{1}{15}$ .

(b) The probability that at most three forms are required is 0.40.

(c) The probability that between two and four forms (inclusive) are required is 0.60.

(d) $P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5$ is not the pmf of y.

Step-by-step explanation:

The random variable Y is defined as the number of forms required of the next applicant.

The probability mass function is defined as:

$P(y) = \left \{ {{ky};\ for \ y=1,2,...5 \atop {0};\ otherwise} \right$

(a)

The sum of all probabilities of an event is 1.

Use this law to compute the value of k.

$\sum P(y) = 1\\k+2k+3k+4k+5k=1\\15k=1\\k=\frac{1}{15}$

Thus, the value of k is $\frac{1}{15}$ .

(b)

Compute the value of P (Y ≤ 3) as follows:

$P(Y\leq 3)=P(Y=1)+P(Y=2)+P(Y=3)\\=\frac{1}{15}+\frac{2}{15}+ \frac{3}{15}\\=\frac{1+2+3}{15}\\ =\frac{6}{15} \\=0.40$

Thus, the probability that at most three forms are required is 0.40.

(c)

Compute the value of P (2 ≤ Y ≤ 4) as follows:

$P(2\leq Y\leq 4)=P(Y=2)+P(Y=3)+P(Y=4)\\=\frac{2}{15}+\frac{3}{15}+\frac{4}{15}\\ =\frac{2+3+4}{15}\\ =\frac{9}{15} \\=0.60$

Thus, the probability that between two and four forms (inclusive) are required is 0.60.

(d)

Now, for $P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5$ to be the pmf of Y it has to satisfy the conditions:

$P(y)=\frac{y^{2}}{50}>0;\ for\ all\ values\ of\ y \\$
$\sum P(y)=1$

Check condition 1:

$y=1:\ P(y)=\frac{y^{2}}{50}=\frac{1}{50}=0.02>0\\y=2:\ P(y)=\frac{y^{2}}{50}=\frac{4}{50}=0.08>0 \\y=3:\ P(y)=\frac{y^{2}}{50}=\frac{9}{50}=0.18>0\\y=4:\ P(y)=\frac{y^{2}}{50}=\frac{16}{50}=0.32>0 \\y=5:\ P(y)=\frac{y^{2}}{50}=\frac{25}{50}=0.50>0$

Condition 1 is fulfilled.

Check condition 2:

$\sum P(y)=0.02+0.08+0.18+0.32+0.50=1.1>1$

Condition 2 is not satisfied.

Thus, $P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5$ is not the pmf of y.

7 0

2 years ago

Please help me! Civics home work super easy

Tomtit [17]

If its easy why dont u do it urself bro

3 0

1 year ago