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r-ruslan [8.4K]
3 years ago
7

An environmental journal discussing current water usage within us reports a 95% confidence interval of (82,88) for the average n

umber of household gallons used per person per day . the study was based on a random sample of 30 individuals throughout the nation. determine the average useage of the 30 individuals sampled
Mathematics
1 answer:
Yuliya22 [10]3 years ago
8 0

Answer:

The average useage of the 30 individuals sampled is 85 gallons per day

Step-by-step explanation:

A confidence interval has two bounds, the lower bound and the upper bound.

A confidence interval is symmetric, which means that the sample mean is the halfway point between these two bounds, that is, the sum of them divided by two.

In this problem, we have that:

Lower bound: 82

Upper bound: 88

Determine the average useage of the 30 individuals sampled

(82 + 88)/2 = 85

The average useage of the 30 individuals sampled is 85 gallons per day

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Match each function with the corresponding function formula when h(x)=5-3x and g(x)=-3+5
Grace [21]

Answer:

k(x) = (3g + 5h)(x) ⇒ (1)

k(x) = (5h - 3g)(x) ⇒ (3)

k(x) = (h - g)(x) ⇒ (2)

k(x) = (g + h)(x) ⇒ (4)

k(x) = (5g + 3h)(x) ⇒ (5)

k(x) = (3h - 5g)(x) ⇒ (6)

Step-by-step explanation:

* To solve this problem we will substitute h(x) and g(x) in k(x) in the

  right column to find the corresponding function formula in the

  left column

∵ h(x) = 5 - 3x

∵ g(x) = -3^x + 5

- Lets start with the right column

# k(x) = (3g + 5h)(x)

∵ g(x) = -3^x + 5

∵ 3g(x) = 3[-3^x + 5] = [3 × -3^x + 3 × 5]

- Lets simplify 3 × -3^x

 take the negative out -(3 × 3^x), and use the rule a^n × a^m = a^(n+m)

∴ -3(3 × 3^x) = -(3^x+1)

∴ 3g(x) = -3^x+1 + 15

∵ h(x) = 5 - 3x

∵ 5h(x) = 5[5 - 3x] = [5 × 5 - 5 × 3x] = 25 - 15x

- Now substitute 3g(x) and 5h(x) in k(x)

∵ k(x) = (3g + 5h)(x)

∴ k(x) = -3^x+1 + 15 + 25 - 15x ⇒ simplify

∴ k(x) = 40 - 3^x+1 - 15x

∴ k(x) = 40 - 3^x+1 - 15x ⇒ k(x) = (3g + 5h)(x)

* k(x) = (3g + 5h)(x) ⇒ (1)

# k(x) = (5h - 3g)(x)

∵ 5h(x) = 25 - 15x

∵ 3g(x) = -3^x+1 + 15

∵ k(x) = (5h - 3g)(x)

∴ k(x) = 25 - 15x - (-3^x+1 + 15) = 25 -15x + 3^x+1 - 15 ⇒ simplify

∴ k(x) = 10 + 3^x+1 - 15x

∴ k(x) = 10 + 3^x+1 - 15x ⇒ k(x) = (5h - 3g)(x)

* k(x) = (5h - 3g)(x) ⇒ (3)

# k(x) = (h - g)(x)

∵ h(x) = 5 - 3x

∵ g(x) = -3^x + 5

∵ k(x) = (h - g)(x)

∴ k(x) = 5 - 3x - (-3^x + 5) = 5 - 3x + 3^x - 5 ⇒ simplify

∴ k(x) = 3^x - 3x

∴ k(x)= 3^x - 3x ⇒ k(x) = (h - g)(x)

* k(x) = (h - g)(x) ⇒ (2)

# k(x) = (g + h)(x)

∵ h(x) = 5 - 3x

∵ g(x) = -3^x + 5

∵ k(x) = (g + h)(x)

∴ k(x) = -3^x + 5 + 5 - 3x ⇒ simplify

∴ k(x) = 10 - 3^x - 3x

∴ k(x)= 10 - 3^x - 3x ⇒ k(x) = (g + h)(x)

* k(x) = (g + h)(x) ⇒ (4)

# k(x) = (5g + 3h)(x)

∵ g(x) = -3^x + 5

∵ 5g(x) = 5[-3^x + 5] = [5 × -3^x + 5 × 5] = 5(-3^x) + 25

∴ 5g(x) = -5(3^x) + 25

∵ h(x) = 5 - 3x

∵ 3h(x) = 3[5 - 3x] = [3 × 5 - 3 × 3x] = 15 - 9x

- Now substitute 5g(x) and 3h(x) in k(x)

∵ k(x) = (5g + 3h)(x)

∴ k(x) = -5(3^x) + 25 + 15 - 9x ⇒ simplify

∴ k(x) = 40 - 5(3^x) - 9x

∴ k(x) = 40 - 5(3^x) - 9x ⇒ k(x) = (5g + 3h)(x)

* k(x) = (5g + 3h)(x) ⇒ (5)

# k(x) = (3h - 5g)(x)

∵ 3h(x) = 15 - 9x

∵ 5g(x) = -5(3^x) + 25

∵ k(x) = (3h - 5g)(x)

∴ k(x) = 15 - 9x - [-5(3^x) + 25] = 15 - 9x + 5(3^x) - 25 ⇒ simplify

∴ k(x) = 5(3^x) - 9x - 10

∴ k(x) = 5(3^x) - 9x - 10 ⇒ k(x) = (3h - 5g)(x)

* k(x) = (3h - 5g)(x) ⇒ (6)

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