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3 years ago

The graph of F(x) = x^2 is shown.

Mathematics

1 answer:

max2010maxim [7]3 years ago

4 0

Answer:

Step-by-step explanation:

Given a function of a parabola (quadratic) in the form f(x) = x^2, we have a translated function as:

g(x) = a(x-b)^2

Where

a is the vertical compression or stretch. If a > 1, it is a vertical stretch and if 0 < a < 1, it is a vertical compression.

b is the horizontal translation b units to the right

The function given is p(x) = 3(x-8)^2

So it means that it is a vertical stretch with a factor 3 and the graph is shifted horizontally 8 units right

the correct answer is B

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3. What is the 9th term in the following sequence? 11, 17, 23, 29, . . . (1 point) 47 53 59 65

KatRina [158]

Since the common difference is 6, we can assume that the sequence is arithmetic, with rule
T(n)=6n+5 where T(1)=11
T(9)=6(9)+5=59

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3 years ago

Is it possible to construct a triangle with side lengths of 15 inches, 17 inches, and 32 inches? Why or why not? (PLEASE HELP ME

dem82 [27]

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3 years ago

The Cubs have won

Naily [24]

First you need to figure out the games the Cubs lost. 11-7=4, so they lost 4 games.

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I hope you find this helpful :)

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3 years ago

Point D is the centroid of Triangle ABC. Find CD and CE.<br> DE = 9

sukhopar [10]

Given:

Point D is the centroid of Triangle ABC and DE = 9.

To find:

The measures of CD and CE.

Solution:

We know that, centroid is the intersection of medians and it divides each median in 2:1.

In triangle ABC, CE is a meaning and centroid D divided CE in 2:1. So,

Let the measures of CD and DE are 2x and x respectively.

DE = 9 (Given)

$x=9$

Now,

$CD=2x$

$CD=2(9)$

$CD=18$

And,

$CE=CD+DE$

$CE=18+9$

$CE=27$

Therefore, the measure of CD is 18 units and the measure of CE is 27 units.

6 0

3 years ago

The local oil changing business is very busy on Saturday mornings and is considering expanding. A national study of similar busi

eimsori [14]

Answer:

$t=\frac{4.2-3.6}{\frac{1.4}{\sqrt{16}}}=1.714$

Reject the null hypothesis if the observed "t" value is less than -2.131 or higher than 2.131

Rejection Zone: $t_{calculated}$ or $t_{calculated}>2.131$

In our case since our calculated value is not on the rejection zone we don't have enough evidence to reject the null hypothesis at 5% of significance.

Step-by-step explanation:

Previous concepts and data given

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=4.2$ represent the sample mean

$s=1.4$ represent the sample standard deviation

n=16 represent the sample selected

$\alpha=0.05$ significance level

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is different from 3.6, the system of hypothesis would be:

Null hypothesis: $\mu = 3.6$

Alternative hypothesis: $\mu \neq 3.6$

If we analyze the size for the sample is < 30 and we 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".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{4.2-3.6}{\frac{1.4}{\sqrt{16}}}=1.714$

Critical values

On this case since we have a bilateral test we need to critical values. We need to use the t distribution with $df=n-1=16-1=15$ degrees of freedom. The value for $\alpha=0.05$ and $\alpha/2=0.025$ so we need to find on the t distribution with 15 degrees of freedom two values that accumulates 0.025 of the ara on each tail. We can use the following excel codes:

"=T.INV(0.025,15)" "=T.INV(1-0.025,15)"

And we got $t_{crit}=\pm 2.131$

So the decision on this case would be:

Reject the null hypothesis if the observed "t" value is less than -2.131 or higher than 2.131

Rejection Zone: $t_{calculated}$ or $t_{calculated}>2.131$

Conclusion

In our case since our calculated value is not on the rejection zone we don't have enough evidence to reject the null hypothesis at 5% of significance.

3 0

3 years ago