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lorasvet [3.4K]
3 years ago
11

I need help on this. Please

Mathematics
1 answer:
serious [3.7K]3 years ago
3 0

Answer: choice C, y = 0.014x+0.85

==============================

Explanation:

Each column of the table represents an (x,y) pair of values

x = number of pages

y = cost

If we look at the first two columns, we see the two points (50,1.55) and (100,2.25). The x value is listed first. Let's compute the slope of the line through these two points

m = (y2-y1)/(x2-x1)

m = (2.25-1.55)/(100-50)

m = 0.7/50

m = 0.014

So far, we see the answer is between A,B or C as they have the slope of 0.014

Use this slope value, and one of the points -- say (x,y) = (50,1.55) -- to find the y intercept b

y = mx+b

y = 0.014x+b .... plug in the slope found earlier

1.55 = 0.014*50+b ... plug in the point (x,y) = (50,1.55)

1.55 = 0.7+b

1.55-0.7 = 0.7+b-0.7 ... subtract 0.7 from both sides

0.85 = b

b = 0.85

With m = 0.014 as the slope and b = 0.85 as the y intercept, we can say that y = mx+b turns into y = 0.014x+0.85. That narrows the answer down to choice C.

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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.

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

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b. P(x ≥ 3) = 0.8050

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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

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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

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