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Tomtit [17]
3 years ago
15

Plz plz plz help on number 6,7,8,9 I would greatly appre and whoever gets it get brainliest

Mathematics
1 answer:
Luden [163]3 years ago
7 0
6: Yes, it is linear. The constant rate of change is \frac{ y_{2}- y_{1} }{ x_{2}- x_{1} }, which you can make \frac{175-100}{10-5}. This simplifies to 15, so the constant rate of change is 15.

7. Yes, a proportional relationship exists because every output value is exactly three times every corresponding input value.

8. Yes, a proportional relationship exists because every output value is exactly \frac{3}{8} times the corresponding input value.

9. Yes, a proportional relationship exists because every y value is 7.5 times every corresponding x value, and if a line were to be drawn through the points, it would go through the origin (0,0).
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Complete Question:

A chemical supply company currently has in stock 100 lb of a certain chemical, which it sells to customers in 5-lb batches. Let X = the number of batches ordered by a randomly chosen customer, and suppose that X has the pmf shown in the table below. Compute E(X) and V(X).

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NOTE: The pmf is shown in the file attached

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Expected number of batches, E(X) = 2.0 batches

Variance of batches, V(X) = 0.8 batches²

Expected number of pounds left, E(Y) = 90 lb

Variance of pounds left, V(Y) = 20 lb²

Step-by-step explanation:

The company has 100 batches and it sells to customers in 5 lb batches. The number of pounds left after each customer is filled will be modeled by the mathematical formula: Y = 100 - 5X

The expected value of X, E(X) can be calculated using the formula:

E(X) = \sum x p(x)

E(X) = (1*0.3) + (2*0.5) + (3*0.1) + (4*0.1)

E(X) = 0.3 + 1 + 0.3 + 0.4

E(X) = 2.0

Expected number of batches, E(X) = 2.0 batches

The variance of X, V(X) can be calculated using the formula:

V(X) = E(X^2) - [E(X)]^2\\ E(X^2) = \sum x^2 p(x)\\  E(X^2) = (1^2 *0.3) + (2^2 * 0.5) + (3^2 * 0.1) + (4^2 * 0.1)\\  E(X^2) = 0.3 + 2 + 0.9 + 1.6\\ E(X^2) = 4.8\\V(X) = 4.8 - 2^2\\V(X) = 0.8

Variance of batches, V(X) = 0.8 batches²

Y = 100 - 5X

E(Y) = E(100 - 5X)

E(Y) = 100 - 5E(X)

E(Y) = 100 - 5(2)

E(Y) = 90 lb

Expected number of pounds left, E(Y) = 90 lb

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V(Y) = V(100 - 5X)

V(Y) = V(100) + V(-5X)

V(100) = 0

V(Y) = (-5)² V(X)

V(Y) = 25 * 0.8

V(Y) = 20

Variance of pounds left, V(Y) = 20 lb²

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