Help please will give brainiest

In segment AC, point B is between A and C. Which of the following is true?

shtirl [24]

Answer:

b

Step-by-step explanation:

4 0

3 years ago

In February, Alec had 1,525 visitors to his website. In March, he had 1,385 visitors to his website. What is the percentage decr

Helga [31]

A proportion box can be used to solve this problem. First, we need to compute the difference or change in the original amount of visitors Alec had (in February) and the new amount of visitors (in March).

1,525 - 1, 385 = 140

We place this amount (140) in the top left of our box, and we place the original amount (1,525) in the bottom left of our box. We are looking for the percentage decrease, so that box in the upper right will stay empty. The number 100 always goes in the bottom right of our percent proportion box.

(The attached image will show an example of how to set this up).

Our last step is to cross multiply and divide.

140 x 100 = 14,000

14,000 ÷ 1,525 = 9.18

So, the percentage decrease of the number of visitors to Alec's website from February to March was 9.18%.

5 0

3 years ago

Greg practiced playing his clarinet for 12 minutes. He has practiced 40% of the amount of time he needs to practiced today. How

frosja888 [35]

He needs to practice 7.2 minutes
he practices for 4.8 minutes in that day

hope this helps

4 0

4 years ago

What statement is true about y=3(x+1/3)^2 + 2/3

faltersainse [42]

Answer:

what are the statements?

Step-by-step explanation:

6 0

3 years ago

Trainees must complete a specific task in less than 2 minutes.Consider the probability density function below for the time ittak

grigory [225]

Answer:

1) P(X<1) = 0.5850

2) P(X>1) = 0.41500

3) P(0.68 < X < 1) = 0.0013

4) E(X) = 0.8867 minute

5) E(X²) = 1.1064

6) Var(X) = 0.3202

Step-by-step explanation:

Probability density function = f(x) = 0.67 - 0.17x (for 0 < x < 2)

1) The probability a trainee will complete the task inless than 1 minutes is given as P(X<1)

And,

P(X<1) = 1 - P(X≥1)

P(X≥1) = ∫²₁ f(x) dx = ∫²₁ (0.67 - 0.17x) dx = [0.67x - 0.085x²]²₁ = [0.67(2) - 0.085(2²)] - [0.67(1) - 0.085(1²)] = (1) - (0.585) = 0.415

P(X<1) = 1 - P(X≥1) = 1 - 0.415 = 0.5850

2) The probability that a trainee will complete the task inmore than 1 minutes is given as P(X>1)

And

P(X>1) = 1 - P(X≤1)

P(X≤1) = ∫¹₀ f(x) dx = ∫¹₀ (0.67 - 0.17x) dx = [0.67x - 0.085x²]¹₀ = (0.67(1) - 0.085(1²)) = 0.67 - 0.085 = 0.585

P(X>1) = 1 - P(X≤1) = 1 - 0.585 = 0.4150

3) The probability it will take a trainee between 0.68minutes and 1 minutes to complete the task. P(0.68 < X < 1)

P(0.68 < X < 1) = P(X<1) - P(X>0.68)

P(X<1) = 0.5850 (from question number 1)

P(X>0.68) = 1 - P(X≤0.68)

P(X≤0.68) = ∫⁰•⁶⁸₀ f(x) dx = ∫⁰•⁶⁸₀ (0.67 - 0.17x) dx = [0.67x - 0.085x²]⁰•⁶⁸₀ = (0.67(0.68) - 0.085(0.68²)) = 0.4556 - 0.039304 = 0.416296 = 0.4163

P(X>0.68) = 1 - P(X≤0.68) = 1 - 0.4163 = 0.5837

P(0.68 < X < 1) = P(X<1) - P(X>0.68) = 0.585 - 0.5837 = 0.001296 = 0.0013 to 4d.p

4) Expected value = E(X) = Σ xᵢpᵢ = ∫²₀ x f(x) dx (that is, a sum of all the products of possible values and their respective probabilities)

∫²₀ x f(x) = ∫²₀ x(0.67 - 0.17x) = ∫²₀ (0.67x - 0.17x²) dx = [0.335x² - 0.0567x³]²₀ = [0.335(2²) - 0.0567(2³)] = 1.34 - 0.4533 = 0.8867 minute.

E(X) = ∫²₀ x f(x) = 0.8867 minute

5) E(X²) = Σ xᵢ²pᵢ = ∫²₀ x² f(x)

∫²₀ x² f(x) = ∫²₀ x²(0.67 - 0.17x) = ∫²₀ (0.67x² - 0.17x³) dx = [0.2233x³ - 0.0425x⁴]²₀ = [0.2233(2)³ - 0.0425(2)⁴] = 1.7864 - 0.68 = 1.1064

6) Variance = Var(X) = Σx²p − μ²

where μ = E(X) = 0.8867 minute

Σx²p = ∫²₀ x² f(x) = 1.1064

Var(X) = 1.1064 - 0.8867² = 0.3202

5 0

3 years ago