1. Which of the following is the first step in the licensing process?

Given:
1 pack = 5 pencils and cardboard.
1 pack should weigh between 60 grams and 95 grams

60g < x < 95g; where x signifies 1 pack.

Cardboard: 15 grams.

95g - 15g = 80g represents the maximum total weight of 5 pencils.
80g / 5 = 16g is the maximum weight for a single pencil.

60g - 15g = 45g is the minimum total weight of 5 pencils.
45g / 5 = 9g is the minimum weight for a single pencil.

9 < x < 16; where x represents a single pencil in the pack.

Answer:

The domain of f(x) indicates x > -2, meaning the range of f-1(x) will be y > -2

The range of f(x) identifies y > -1, thus the domain of f-1(x) is x > -1

Step-by-step explanation:

Answer:

Please review the explanation

Detailed explanation:

Based on the data provided:

Time, t (hours)__1.0__1.5__2.0__2.5__3.0

C(t) (mg/mL)__0.35_0.26_0.20_0.14_0.09

Average change of C with respect to t over the interval:

(i) [1.0, 2.0] (ii) [1.5, 2.0] (iii) [2.0, 2.5] (iv) [2.0, 3.0]

Average change is calculated as ( change in C / change in t) = C2 - C1 / t2 - t1

1) [1.0, 2.0]

C at t = 1 = 0.35; C at t = 2 = 0.20

(0.20 - 0.35) / (2 - 1) = - 0.15 / 1 = - 0.15 mg/mL hr

11) [1.5, 2.0]

(0.20 - 0.26) / (2.0 - 1.5) = - 0.06 / 0.5 = - 0.12 mg/mL hr

111) [2.0, 2.5]

(0.14 - 0.20) / (2.5 - 2.0) = - 0.06 / 0.5 = - 0.12mg/mL hr

iv) [2.0, 3.0]

(0.09 - 0.20) / (3.0 - 2.0) = - 0.11 / 1.0 = - 0.11 mg/mL hr

Answer:

Jada's mistake was that he multiplied the equation by (-9/4) instead of using 108 to achieve a coefficient of one for x. He ought to have multiplied by 108

Step-by-step explanation:

Jada tackled the equation

-4/9 = x/108

following these steps:

-4/9 = x/108

(-4/9)(-9/4) = (x/108)(-9/4)

x = -1/48

The error was multiplying by (-4/9) when multiplying by 108 was the correct approach.

If he had multiplied by 108, it would result in

(-4/9)(108) = (x/108)(108)

-48 = x

x = -48

Thus, the correct answer should have been

x = -48

rather than

x = -1/48

To resolve this question, simply tally the probabilities of the options. The likelihood of a pitch not crossing the plate being a strike is non-existent; thus, P(A | D) = 0. This is true as it calculates to 0/0+20= 0. The likelihood of a pitch not over the plate being a ball is complete; therefore, P(B | D) = 1. This holds true as it equals 20/20+0= 1. The likelihood of a pitch over the plate being a strike is 10:15. It’s incomplete but seems correct; it can be calculated as 10/10+5= 10/15 = 2/3. The probability for a pitch over the plate labeled a ball is 5:10; therefore, P(B | C) = 0.5.