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2 years ago

The graph of f(x)=(c) is translated 6 units to the right and 2 units up to form a new function

Mathematics

1 answer:

jasenka [17]2 years ago

5 0

Answer:

Step-by-step explanation:

My opinion is that you meant "f(x) = c." This would be a constant function whose graph is a horizontal line, y = c.

If this graph were translated 6 units to the right, the function would not change.

However, if this function f(x) = c were translated 2 units up, the new function would be g(x) = c + 2.

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A researcher compares the effectiveness of two different instructional methods for teaching electronics. A sample of 138 student

yan [13]

Answer:

The 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 is (-8.04, 0.84).

Step-by-step explanation:

The (1 - α)% confidence interval for the difference between population means is:

$CI=(\bar x_{1}-\bar x_{2})\pm z_{\alpha/2}\times \sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}$

The information provided is as follows:

$n_{1}= 138\\n_{2}=156\\\bar x_{1}=61\\\bar x_{2}=64.6\\\sigma_{1}=18.53\\\sigma_{2}=13.43$

The critical value of z for 98% confidence level is,

$z_{\alpha/2}=z_{0.02/2}=2.326$

Compute the 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 as follows:

$CI=(\bar x_{1}-\bar x_{2})\pm z_{\alpha/2}\times \sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}$

$=(61-64.6)\pm 2.326\times\sqrt{\frac{(18.53)^{2}}{138}+\frac{(13.43)^{2}}{156}}\\\\=-3.6\pm 4.4404\\\\=(-8.0404, 0.8404)\\\\\approx (-8.04, 0.84)$

Thus, the 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 is (-8.04, 0.84).

5 0

2 years ago

Phil is 2 years older than Andrew. Andrew is 6 years old. The sum of their ages is 22 years. How old is Bob?

KatRina [158]

Answer:

he would be 14

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Let g(x)=2x and h(x)=x^2+4 find the value 1.) (g*h)(0) 2.) (h*g) (a)

pentagon [3]

(HoG)(x) = (2x)2 + 4 simply because HoG(x) is actually H(G(x)). So where ever there was an x in H(x) we substitute our value of G(x). Now the only thing left is to put in the 1. so out answer is (2(1))2 + 4 = 8Only in the 6th grade ; )

3 0

2 years ago

Read 2 more answers

Chris plans to buy a car that depreciates in value at a rate of 10% per year. If the car initially costs $32,000. How much will

VLD [36.1K]

The car would be worth $16,000 after 5 years.

5 0

2 years ago

Susan threw a softball 42 years on her first try and 51 1/3 yard on her second try. How much farther did she throw the softball

Andru [333]

Answer:

28 feet farther than 1st ball.

Step-by-step explanation:

We have been given that Susan threw a softball 42 yards on her first try and $51\frac{1}{3}$ yard on her second try.

To find second ball is how much farther from the 1st ball, we will subtract 42 yards from $51\frac{1}{3}$ yards.

$\text{The second ball is farther from 1st ball}=51\frac{1}{3}\text{ yards}-42\text{ yards}$

$\text{The second ball is farther from 1st ball}=\frac{154}{3}\text{ yards}-42\text{ yards}$

Let us have a common denominator.

$\text{The second ball is farther from 1st ball}=\frac{154}{3}\text{ yards}-\frac{42*3}{3}\text{ yards}$

$\text{The second ball is farther from 1st ball}=\frac{154}{3}\text{ yards}-\frac{126}{3}\text{ yards}$

$\text{The second ball is farther from 1st ball}=\frac{154-126}{3}\text{ yards}$

$\text{The second ball is farther from 1st ball}=\frac{28}{3}\text{ yards}$

$\text{1 yard}=3\text{ feet}$

$\frac{28}{3}\text{ yards}=\frac{28}{3}\times 3\text{ feet}$

$\frac{28}{3}\text{ yards}=28\text{ feet}$

Therefore, Susan thrown the second ball 28 feet farther from the 1st ball.

7 0

3 years ago