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2 years ago

What is the horizontal and vertical shift for the absolute value function below f(x)=3[x+9]+6

Mathematics

1 answer:

salantis [7]2 years ago

4 0

We will see that we have a shift of 9 units to the left and 6 units upwards

<h3>How to identify the translations?</h3>

For any given function H(x), we define a vertical shift of N units as:

g(x) = H(x) + N.

If N > 0, the shift is upwards.
If N < 0, the shift is downwards.

While a horizontal shift is written as:

g(x) = H(x + N).

If N > 0, the shift is to the left.
if N < 0, the shift is to the right.

Here we know that:

f(x) = 3*|x + 9| + 6

If the original function was:

H(x) = 3*|x|

Then we can write:

f(x) = H(x + 9) + 6

So we have a shift of 9 units to the left and 6 units upwards.

If you want to learn more about translations:

brainly.com/question/24850937

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2 years ago

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Answer:

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Step-by-step explanation:

We are given that Breyers is a major producer of ice cream and would like to test if the average American consumes more than 17 ounces of ice cream per month.

A random sample of 25 Americans was found to consume an average of 19 ounces of ice cream last month. The standard deviation for this sample was 5 ounces.

Let $\mu$ = average ounces of ice cream consumed by American per month

So, Null Hypothesis, $H_0$ : $\mu \leq$ 17 ounces {means that the average American consumes less than or equal to 17 ounces of ice cream per month}

Alternate Hypothesis, $H_A$ : $\mu$ > 17 ounces {means that the average American consumes more than 17 ounces of ice cream per month}

The test statistics that will be used here is One-sample t test statistics as we don't know about population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample average = 19 ounces

s = sample standard deviation = 5 ounces

n = sample of Americans = 25

So, test statistics = $\frac{19-17}{\frac{5}{\sqrt{25} } }$ ~ $t_2_4$

= 2

The value of the test statistics is 2.

Now at 0.025 significance level, the t table gives critical value of 2.06 at 24 degree of freedom for right-tailed test. Since our test statistics is less than the critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

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