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3 years ago

what is the function rule for the transformed with vertical stretch by a factor of 2 and translated down 6

1 answer:

5 0

Answer: Thus, the equation of a function stretched vertically by a factor of 2 and then shifted 3 units up is y = 2f (x) + 6, and the equation of a function stretched horizontally by a factor of 2 and then shifted 6 units left is y = f ( (x - 6)) = f ( x - ). Example: f (x) = 2x2.

Step-by-step explanation:

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Hello, this is math and I need help ❤️ pls and ty ONLY SERIOUS ANSWERS

Greeley [361]

The bicycle with a 30-inches diameter rolls approximately 28.8 more than a tire with the diameter with 21-inches.

<u><em>How to solve:</em></u>

First use the formula: C = 3.14d
Replace the d value with 30 and solve the equation:

C = 3.14 x 30 ⇒ 94.2477796077

Then solve the equation again but 21 as the d value:

C = 3.14 x 21 ⇒ 65.9734457254

Then subtract 65.9734457254 from 94.2477796077 to get the approx. answer of 28.8

3 0

3 years ago

Someone please help me with my homework!! plzzzzz

vodomira [7]

Y=21

Step-by-step explanation:

8 0

3 years ago

What is the range of the function shown on the graph?

klasskru [66]

I need to see the graph.

6 0

3 years ago

Mary mixes white paint and blue paint in the ratio 2:5.

sukhopar [10]

Answer:Add 44 liters blue

Step-by-step explanation:Mary has 20 liters of paint mixed white:blue in the ratio 2:3. So two fifths of that paint is white and three fifths is blue. So that's 8 liters white, 12 liters blue.

We want to add paint to get white:blue = 1:7. Clearly we need to add blue. We have our 8 liters of white, so we need a total of 7(8)=56 liters of blue. We already have 12, so we have to add 56-12=44 liters of blue paint:

6 0

3 years ago

Young people may feel they are carrying the weight of the world on their shoulders when in reality they are too often carrying a

Travka [436]

Answer:

A. [12.92;14.97]lb

B. [13.25;16.26]%

C. check explanation

Step-by-step explanation:

Hello!

To properly resolve any statistical problem you need to first identify your study variables and summarize the given data.

Study variable:

X: "backpack weight of a sixth-grader" (lb)

sample n= 131

sample mean: x[bar]= 13.83 lb

sample standard deviation: S= 5.05 lb

Now you need to calculate a CI for the population mean of the backpack weight of sixth graders (μ) with a confidence level of 99%

Since you need to estimate the population mean, and you have a sample large enough (n≥30), although you don't know the distribution of this population, you can approximate the distribution of the sample mean to normal (applying CLT) and use the statistic Z= (x[bar] - μ)/(S/√n) ≈ N(0;1)

The formula for the CI is

x[bar] ± $Z_{\alpha/2}$ * (S/√n)

The confidence level to use is 1-α = 0.99

α = 0.01 ⇒ α/2 = 0.005

$Z_{\alpha/2}$ ⇒ $Z_{\0.005}$ = -2.58

Now you calculate the interval

13.83 ± 2.58 * 5.05/√131 ⇒ [12.92;14.97]lb

So with a confidence level of 99%, you'll expect that the real mean for the backpack weight of sixth graders is contained by the interval [12.92;14.97]lb

It seems that for this item the information was measured from the same sample taken in item A.

Study variable

X: "Backpack weight as percentage of body weight" (lb)

Assuming is the same sample: n= 131

It was calculated the 95% CI [13.62;15.89]lb

And you need to calculate a 99% CI

For this item, you use the same statistic as before

x[bar] ± $Z_{\alpha/2}$ * (S/√n)

To calculate the asked interval you need to know the confidence level, the sample mean, sample standard deviation and sample size.

To know the sample mean and standard deviation for the backpack weight as percentage of body weight, you'll need to deduce it from the given interval.

Since this interval is for the population mean, it is constructed arround the sample mean. That means, that the sample mean is at the center of the interval.

x[bar] = (Li+Ls)/2 ⇒ x[bar] = (13.62+15.89)/2 = 14.76%

The standard deviation needs a little more calculation,

amplitude a= Ls - Li ⇒ a= 15.89-13.62= 2.27

semiamplitude d= a/2 ⇒ d=2.27/2= 1.135lb ≅ 1.14

$Z_{\1-alpha/2}$ ⇒ $Z_{\0.975}$ = 1.96

d= $Z_{\alpha/2}$ * (S/√n) ⇒ 1.14= 1.96 * (S/√131)

⇒ S= 6.657% ≅ 6.66%

Now the new CI has a confidence level of 99%

The Z value you need to use is $Z_{\0.995}$ = 2.58

x[bar] ± $Z_{\alpha/2}$ * (S/√n) ⇒ 14.76 ± 2.58 * (6.66/√131)

⇒ [13.25;16.26]%

With a confidence level of 99% you can expect that the interval [13.25;16.26]% contains the population mean of the backpack weight as percentage of body weight of sixth graders.

"The American Academy of Orthopedic Surgeons recommends that backpack weight is at most 10% of body weight." As you can see in the calculated interval [13.25;16.26]%, the weight carried by the sixth graders is above the recommended backpack weight as percentage of body weight. This means that the sixth graders carry more weight than recommended.

I hope you have a SUPER day!

4 0

3 years ago