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2 years ago

Refer to the equation x - 5y = 20.

Mathematics

1 answer:

Rainbow [258]2 years ago

5 0

Answer:

Kindly check explanation

Step-by-step explanation:

Given the equation :

X - 5y = 20

x - 20 = 5y

(x - 20) / 5 = y

x = - 5

y = (-5 - 20) / 5 = -5

x = 0

y = (0 - 20) / 5 = - 4

x = 5

y = (5 - 20) / 5 = - 3

x = 10

y = (10 - 20) / 5 = - 2

X : _____ - 5 ___ 0 ____ 5 ____10

y : _____ - 5 ___- 4 ___ - 3 ___ - 2

Kindly make use of a graph for better clarity and accuracy

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2 years ago

In a MBS first year class, there are three sections each including 20 students. In the first section, there are 10 boys and 10 g

KIM [24]

Answer:

$3.52 \times 10^{-9} = 3.52 \times 10^{-7}\%$ probability that all the 15 students selected are girls

Step-by-step explanation:

The selection is from a sample without replacement, so we use the hypergeometric distribution to solve this question.

Hypergeometric distribution:

The probability of x sucesses is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

All girls from the first group:

20 students, so $N = 20$

10 girls, so $k = 10$

5 students will be selected, so $n = 5$

We want all of them to be girls, so we find P(X = 5).

$P_1 = P(X = 5) = h(5,20,5,10) = \frac{C_{10,5}*C_{10,5}}{C_{20,5}} = 0.0163$

All girls from the second group:

20 students, so $N = 20$

5 girls, so $k = 5$

5 students will be selected, so $n = 5$

We want all of them to be girls, so we find P(X = 5).

$P_2 = P(X = 5) = h(5,20,5,5) = \frac{C_{5,5}*C_{15,5}}{C_{20,5}} = 0.00006$

All girls from the third group:

20 students, so $N = 20$

8 girls, so $k = 8$

5 students will be selected, so $n = 5$

We want all of them to be girls, so we find P(X = 5).

$P_3 = P(X = 5) = h(5,20,5,8) = \frac{C_{8,5}*C_{12,5}}{C_{20,5}} = 0.0036$

All 15 students are girls:

Groups are independent, so we multiply the probabilities:

$P = P_1*P_2*P_3 = 0.0163*0.00006*0.0036 = 3.52 \times 10^{-9}$

$3.52 \times 10^{-9} = 3.52 \times 10^{-7}\%$ probability that all the 15 students selected are girls

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2 years ago

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Sliva [168]

Answer:

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Step-by-step explanation:

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$280 = Pe^0.019(12)

$280 = Pe^0.228, then evaluate the solution by isolating the variable and dividing each side by the factors which don't contain the variable.

P = 222.91479275

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2 years ago