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3 years ago

Describe the transformation from the graph of f(x) =ax^2 to the graph of g(x) = a(x-h)^2+k

Mathematics

1 answer:

MrRa [10]3 years ago

8 0

Answer: Step-by-step explanation:

f(x)=ax^2→g(x)=a(x-h)^2+k

The horizontal shift depends on the value of h. The horizontal shift is described as:

g(x)=f(x+h) - The graph is shifted to the left h units.

g(x)=f(x−h) - The graph is shifted to the right h units.

Hence, in the question above, Horizontal Shift: Right h Units

The vertical shift depends on the value of k. The vertical shift is described as:

g(x)=f(x)+k - The graph is shifted up k units.

g(x)=f(x)−k - The graph is shifted down k units.

Therefore, in the question above, Vertical Shift: Up k Units

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max2010maxim [7]

Answer:

3.x=20

Step-by-step explanation:

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3 years ago

The length of a rectangular is 5/2 unites greater then twice its width if it’s width w which expression gives the perimeter of t

strojnjashka [21]

If width is w the length can be represented as 5/2+2w.

Formula for perimeter is 2l + 2w

So it would be 2w + 2(5/2 + 2w)

OR 2w + 5 + 4w

OR 6w + 5

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3 years ago

Solve the inequality.<br><br> y – 2 > 3

wlad13 [49]

Answer:

Step-by-step explanation:

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3 years ago

In 1980, the median age of the U.S. population was 30.0; in 2000, the median age was 35.3.

miss Akunina [59]

Answer:

y = 0.265x - 494.7

Step-by-step explanation:

Let median age be represent by 'a' and time be represent by 't'

In 1980, median age is given 30

which means that

a₁ = 30

t₁ = 1980

In 2000, the median age is given 35.3

which means that.

a₂ = 35.3

t₂ = 2000

The slope 'm' of the linear equation can be found by:

m = (a₂ - a₁) /(t₂ - t₁)

m = (35.3 - 30)/(2000-1980)

m = 0.265

General form of linear equation is given by:

y = mx + c

y = 0.265x +c

Substitute point (1980,30) in the equation.

30 = 0.265(1980) + c

c = -494.7

Hence the the linear equation can be written as:

y = mx + c

y = 0.265x - 494.7

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3 years ago

A random sample of 25 ACME employees showed the average number of vacation days taken during the year is 18.3 days with a standa

Norma-Jean [14]

Answer:

a) Null hypothesis: $\mu \leq 15$

Alternative hypothesis: $\mu > 15$

b) $df=n-1=25-1=24$

For this case the p value is given $p_v = 0.0392$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is higher than 15 at 5% of signficance.

c) Type I error, also known as a “false positive” is the error of rejecting a null hypothesis when it is actually true. Can be interpreted as the error of no reject an alternative hypothesis when the results can be attributed not to the reality.

So for this case a type I of error would be reject the hypothesis that the true mean is less or equal than 15 and is actually true.

Step-by-step explanation:

Data given and notation

$\bar X=18.3$ represent the sample mean

$s=3.72$ represent the sample standard deviation

$n=25$ sample size

$\mu_o =15$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

Part a: State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean for vacation days is higher than 15, the system of hypothesis would be:

Null hypothesis: $\mu \leq 15$

Alternative hypothesis: $\mu > 15$

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Part b: P-value and conclusion

The first step is calculate the degrees of freedom, on this case:

$df=n-1=25-1=24$

For this case the p value is given $p_v = 0.0392$

Conclusion

Part c

Type I error, also known as a “false positive” is the error of rejecting a null hypothesis when it is actually true. Can be interpreted as the error of no reject an alternative hypothesis when the results can be attributed not to the reality.

So for this case a type I of error would be reject the hypothesis that the true mean is less or equal than 15 and is actually true.

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3 years ago