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3 years ago

Which equations and/or functions represent the graphed line? Check all that apply.

Mathematics

2 answers:

8090 [49]3 years ago

3 0

The picture in the attached figure

step 1

find the equation of the line of the graph

let
A (-4,0) B(0,2)

slope m=(y2-y1)/(x2-x1)------> m=(2-0)/(0+4)----> m=1/2----> m=0.5

with m=0.5 and point B (0,2)
y-y1=m*(x-x1)-----> y-2=(0.5)*(x-0)-----> y=0.5x+2

step 2
find the solution

a) f(x) = 0.2x - 4----> the function not represent the graphed line
b) f(x) = 0.5x + 2----> the function represent the graphed line
c) f(x) = x + 2--> the function not represent the graphed line
d) y – 3 = (x – 2)---> y=x+1---> the function not represent the graphed line
e) y – 1 = 0.5(x + 2)-----> y=0.5x+1+1---> y=0.5x+2-----> the function represent the graphed line

therefore

the answer is
f(x) = 0.5x + 2
and
y – 1 = 0.5(x + 2)

alisha [4.7K]3 years ago

3 0

Answer:

the answer is

f(x) = 0.5x + 2

and

y – 1 = 0.5(x + 2)

Step-by-step explanation:

well this is based on what i think so hope it helps. XD

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Jesse sold cupcakes and cookies yesterday. Each cupcake sold for $1.50 and each cookie sold for $0.25. At the end of the day, Je

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Answer: 19 cupcakes and 37 cookies were sold.

Step-by-step explanation:

Let x represent the number of cupcakes that Jesse sold yesterday.

Let y represent the number of cookies that Jesse sold yesterday.

If he sold 56 cupcakes and cookies combined, it means that

x + y = 56

Each cupcake sold for $1.50 and each cookie sold for $0.25. At the end of the day, Jesse had sold $37.75 worth of cookies and cupcakes. This means that

1.5x + 0.25y = 37.75 - - - - - - - - - - -1

Substituting x = 56 - y into equation 1, it becomes

1.5(56 - y) + 0.25y = 37.75

84 - 1.5y + 0.25y = 37.75

- 1.5y + 0.25y = 37.75 - 84

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y = - 46.25/ - 1.25

y = 37

x = 56 - y = 56 - 37

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2 years ago

The following data represent the pH of rain for a random sample of 12 rain dates in Tucker County, West Virginia. A normal proba

Sphinxa [80]

Answer:

Step-by-step explanation:

n = 12

Mean = (4.58 + 5.72 + 4.77 + 4.76 + 5.19 + 5.05 + 4.80 + 4.77 + 4.75 + 5.02 + 4.74 + 4.56)/12 = 4.8925

Standard deviation = √(summation(x - mean)²/n

Summation(x - mean)² = (4.58 - 4.8925)^2 + (5.72 - 4.8925)^2 + (4.77 - 4.8925)^2 + (4.76 - 4.8925)^2 + (5.19 - 4.8925)^2 + (5.05 - 4.8925)^2 + (4.80 - 4.8925)^2 + (4.77 - 4.8925)^2 + (4.75 - 4.8925)^2 + (5.02 - 4.8925)^2 + (4.74 - 4.8925)^2 + (4.56 - 4.8925)^2 = 1.122225

Standard deviation = √(1.122225/12

s = 0.31

a) Point estimate = sample mean = 4.8925

Confidence interval is written in the form,

(Sample mean - margin of error, sample mean + margin of error)

Margin of error = z × s/√n

Where

From the information given, the population standard deviation is unknown and the sample size is small, hence, we would use the t distribution to find the z score

In order to use the t distribution, we would determine the degree of freedom, df for the sample.

df = n - 1 = 12 - 1 = 11

b) Since confidence level = 95% = 0.95, α = 1 - CL = 1 – 0.95 = 0.05

α/2 = 0.05/2 = 0.025

the area to the right of z0.025 is 0.025 and the area to the left of z0.025 is 1 - 0.025 = 0.975

Looking at the t distribution table,

z = 2.201

Margin of error = 2.201 × 0.31/√12

= 0.197

95% confidence interval = 4.8925 ± 0.197

Upper limit = 4.8925 + 0.197 = 5.0895

Lower limit = 4.8925 - 0.197 = 4.6955

We are 95% confident that the population mean of the rain water ph lies between 4.6955 and 5.0895

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Margin of error = 3.106 × 0.31/√12

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99% confidence interval = 4.8925 ± 0.278

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Lower limit = 4.8925 - 0.278 = 4.6145

We are 99% confident that the population mean of the rain water ph lies between 4.6145 and 5.1705

d) The interval gets wider as the confidence level is increased. This is logical since the test score is higher for 99% and therefore, increases the range of values. Since we want to be more confident, the range of values must be extended.

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Answer:

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