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nasty-shy [4]
4 years ago
12

Solve y=1/4x-1 if the domain is {-4,-2,0,2,4}.

Mathematics
1 answer:
kipiarov [429]4 years ago
5 0

Allowed values for y: -2,-\frac{3}{2},-1,-\frac{1}{2},0

Step-by-step explanation:

The domain of a function f(x) represents the set of values for which the function is defined. In other words, it represents the values of x that are allowed.

The function in this problem is:

y=\frac{1}{4}x-1

And the domain is

D: {-4,-2,0,2,4}

This means that these are the only allowed values for x. We can find the range of the functions (the possible values for y) simply by substituting these values of x into the function. We get:

For x = -4,

y=\frac{1}{4}(-4)-1=-1-1=-2

For x = -2,

y=\frac{1}{4}(-2)-1=-\frac{1}{2}-1=-\frac{3}{2}

For x = 0,

y=\frac{1}{4}(0)-1=-1

For x = 2,

y=\frac{1}{4}(2)-1=\frac{1}{2}-1=-\frac{1}{2}

For x = 4,

y=\frac{1}{4}(4)-1=1-1=0

Therefore, the allowed values for y are:

-2,-\frac{3}{2},-1,-\frac{1}{2},0

Learn more about domain and range:

brainly.com/question/7128279

brainly.com/question/9607945

brainly.com/question/1485338

#LearnwithBrainly

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A quality analyst of a tennis racquet manufacturing plant investigates if the length of a junior's tennis racquet conforms to th
Burka [1]

Answer:

Confidence interval : 21.506 to 24.493

Step-by-step explanation:

A quality analyst selects twenty racquets and obtains the following lengths:

21, 25, 23, 22, 24, 21, 25, 21, 23, 26, 21, 24, 22, 24, 23, 21, 21, 26, 23, 24

So, sample size = n =20

Now we are supposed to find Construct a 99.9% confidence interval for the mean length of all the junior's tennis racquets manufactured at this plant.

Since n < 30

So we will use t-distribution

Confidence level = 99.9%

Significance level = α = 0.001

Now calculate the sample mean

X=21, 25, 23, 22, 24, 21, 25, 21, 23, 26, 21, 24, 22, 24, 23, 21, 21, 26, 23, 24

Sample mean = \bar{x}=\frac{\sum x}{n}

Sample mean = \bar{x}=\frac{21+25+23+22+24+21+25+21+23+ 26+ 21+24+22+ 24+23+21+ 21+ 26+23+ 24}{20}

Sample mean = \bar{x}=23

Sample standard deviation = \sqrt{\frac{\sum(x-\bar{x})^2}{n-1}}

Sample standard deviation = \sqrt{\frac{(21-23)^2+(25-23)^2+(23-23)^2+(22-23)^2+(24-23)^2+(21-23)^2+(25-23)^2+(21-23)^2+(23-23)^2+(26-23)^2+(21-23)^2+(24-23)^2+(22-23)^2+(24-23)^2+(23-23)^2+(21-23)^2+(21-23)^2+(26-23)^2+(23-23)^2+(24-23)^2}{20-1}}

Sample standard deviation= s = 1.72

Degree of freedom = n-1 = 20-1 -19

Critical value of t using the t-distribution table t_{\frac{\alpha}{2} = 3.883

Formula of confidence interval : \bar{x} \pm t_{\frac{\alpha}{2}} \times \frac{s}{\sqrt{n}}

Substitute the values in the formula

Confidence interval : 23 \pm 1.73 \times \frac{1.72}{\sqrt{20}}

Confidence interval : 23 -3.883 \times \frac{1.72}{\sqrt{20}} to 23 + 3.883 \times \frac{1.72}{\sqrt{20}}

Confidence interval : 21.506 to 24.493

Hence Confidence interval : 21.506 to 24.493

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4 years ago
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