What is the greatest common factor of 28 and 56

What is the simplified answer? 13x + 6 - 8x -2

mario62 [17]

5x+4 u add -8x and 13x and then -2 +6

5 0

3 years ago

A car is driving down a long Steep Hill the elevation above sea level in feet of the car when it is D miles from the top of the

amid [387]

Answer:

This task would be especially well-suited for instructional purposes. Students will benefit from a class discussion about the slope, y-intercept, x-intercept, and implications of the restricted domain for interpreting more precisely what the equation is modeling.

One potential confusion students may have follows from the subtle difference between what the car is doing and the idea of slope as the ratio between the change in vertical distance on the graph and the change in horizontal distance on the graph. Because the car is traveling one mile on a down-hill slope, the situation could be represented as a right triangle with a hypotenuse of 5,280 ft and a leg of 250 ft; using the Pythagorean Theorem they would find that the other leg is approximately 5,274 ft. Following through on this interpretation, a student might conclude that the car travels a horizontal distance of approximately 5,274 ft for every 250 ft in vertical distance and arrive at a slope of approximately -0.047. While this is, in some sense, the slope of the hill, it is not the slope of the function as described. This interpretation yields numbers that are very close to the situation described in the task, yet conceptually different since the distance traveled by the car would now be expressed in terms of horizontal distance traveled as opposed to distance along the slope of the hill to compute the elevation. If students do indeed pursue this line of reasoning, the task provides an opportunity to compare and contrast the graph of the function and what it represents with a drawing of the hill and the vertical and horizontal distances traversed with each mile down the slope.

Step-by-step explanation:

5 0

2 years ago

Market-share-analysis company Net Applications monitors and reports on Internet browser usage. According to Net Applications, in

arsen [322]

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

Market-share-analysis company Net Applications monitors and reports on Internet browser usage. According to Net Applications, in the summer of 2014, Google's Chrome browser exceeded a 20% market share for the first time, with a 20.37% share of the browser market (Forbes website, December ). For a randomly selected group of 20 Internet browser users, answer the following questions.

a. Compute the probability that exactly 8 of the 20 Internet browser users use Chrome as their Internet browser (to 4 decimals). For this question, if you compute the probability manually, make sure to carry at least six decimal digits in your calculations.

b. Compute the probability that at least 3 of the 20 Internet browser users use Chrome as their Internet browser (to 4 decimals).

c. For the sample of Internet browser users, compute the expected number of Chrome users (to 3 decimals).

d. For the sample of Internet browser users, compute the variance and standard deviation for the number of Chrome users (to 3 decimals).

Answer:

a. P(x = 8) = 0.024273

b. P(x ≥ 3) = 0.8050

c. E(x) = 4.074

d. var(x) = 3.244 and SD = 1.801

Step-by-step explanation:

The given problem can be solved using the binomial distribution

P(x) = ⁿCₓ pˣ (1 - p)ⁿ⁻ˣ

Where n is the number of trials, x is the variable of interest and p is the probability of success.

For the given problem,

Probability of success = p = 20.37% = 0.2037

Probability of failure = q = 1 - p = 1- 0.2037 = 0.7963

Number of trials = n = 20

a. Compute the probability that exactly 8 of the 20 Internet browser users use Chrome as their Internet browser.

In this case, we have x = 8

P(x = 8) = ²⁰C₈×0.2037⁸×(1 - 0.2037)²⁰⁻⁸

P(x = 8) = 125970×0.2037⁸×0.7963¹²

P(x = 8) = 0.024273

b. Compute the probability that at least 3 of the 20 Internet browser users use Chrome as their Internet browser (to 4 decimals).

P(x ≥ 3) = 1 - P(x < 2)

But we know that

P(x < 2) = P(x = 0) + P(x = 1) + P(x = 2)

So,

P(x ≥ 3) = 1 - [ P(x = 0) + P(x = 1) + P(x = 2) ]

For P(x = 0):

Here we have x = 0, n = 20 and p = 0.2037

P(x = 0) = ²⁰C₀×0.2037⁰×(1 - 0.2037)²⁰⁻⁰

P(x = 0) = 0.0105

For P(x = 1):

Here we have x = 1, n = 20 and p = 0.2037

P(x = 1) = ²⁰C₁×0.2037¹×(1 - 0.2037)²⁰⁻¹

P(x = 1) = 0.0538

For P(x = 2):

Here we have x = 2, n = 20 and p = 0.2037

P(x = 2) = ²⁰C₂×0.2037²×(1 - 0.2037)²⁰⁻²

P(x = 2) = 0.1307

Finally,

P(x ≥ 3) = 1 - [P(x = 0) + P(x = 1) + P(x = 2)]

P(x ≥ 3) = 1 - [ 0.0105 + 0.0538 + 0.1307]

P(x ≥ 3) = 1 - [0.1950]

P(x ≥ 3) = 0.8050

c. For the sample of Internet browser users, compute the expected number of Chrome users (to 3 decimals).

The expected number of Chrome users is given by

E(x) = n×p

Where n is the number of trials and p is the probability of success

E(x) = 20×0.2037

E(x) = 4.074

d. For the sample of Internet browser users, compute the variance and standard deviation for the number of Chrome users (to 3 decimals).

The variance for the number of Chrome users is given by

var(x) = n×p×q

Where n is the number of trials and p is the probability of success and q is the probability of failure.

var(x) = 20×0.2037×0.7963

var(x) = 3.244

The standard deviation for the number of Chrome users is given by

SD = √(n×p×q)

SD = √var(x)

SD = √(3.244)

SD = 1.801

5 0

3 years ago

Write the standard form of the line that passes through the given points (6,1) and (5,4)

Serggg [28]

The standard form:

$Ax+By=C$

The point-slope form:

$y-y_1=m(x-x_1)$

m - slope

The formula of a slope:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

We have the points (6, 1) and (5, 4). Substitute:

$m=\dfrac{4-1}{5-6}=\dfrac{3}{-1}=-3$

$y-1=-3(x-1)$ use distributive property

$y-1=-3x+3$ add 1 to both sides

$y=-3x+4$ add 3x to both sides

$\boxed{3x+y=4}$

7 0

3 years ago

The manufacturer of official NFL footballs uses a machine to inflate its new balls to a pressure of 13.5 pounds. When the machin

Dmitriy789 [7]

Answer:

106

Step-by-step explanation:

Given that:

Mean (m) = 13.5 pounds

Error margin = 0.025

α = 99%

Variation interval = range = 13.3 to 13.7

Standard deviation (σ) estimate from range :

Range / 4

(13.7 - 13.3) /4

= 0.4 / 4

= 0.1

Zcritical at 99% = 2.576

Using the formula :

Sample size (n) = [ (Zcritical * σ) / M. E]^2

n = [(2.576 * 0.1) / 0.025]^2

n = (0.2576 / 0.025)

n = 10.304^2

n = 106.172

n = 107 samples

5 0

3 years ago