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3 years ago

12

For what values of a and m does f(x) have a horizontal asymptote at y = 2 and a vertical asymptote at x = 1? f(x)=2x^m/x+a {a =

–1, m = 1}

1 answer:

iVinArrow [24]3 years ago

3 0

Answer:

See below

Step-by-step explanation:

$f(x)=\frac{2x^{m} }{x+a}$

$f(x)=\frac{2x^{1} }{x-1}$

If x=1, then 1-1=0, which implies a vertical asymptote at x=1 since dividing by 0 is undefined.

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Frank’s dog eats 6 ounces of dog food each day. Frank bought a 42-pound bag of dog food. How many 6-ounce servings are in 42-pou

iVinArrow [24]

So it’ll be 42 divided by 6 since u want to know how many servings of 6 ounces can be served from a 42 pound bag
So 42/6= 7

5 0

2 years ago

College freshmen took a psychology exam. If the mean is 80, the SD is 10, and the scores have normal distribution, what percent

gayaneshka [121]

Answer:

(a) 2%

Step-by-step explanation:

A score of 60 is 2 standard deviations below the mean (80 -2·10 = 60). The empirical rule tells you that 95% of the distribution is within 2 standard deviations of the mean. The remaining 5% is split evenly between those that are more than 2 SD above the mean and those that are more than 2 SD below the mean.

That tells you the percentage who failed is about 2.5%. The best answer choice is 2%.

_____

<em>Additional comment</em>

The empirical rule is a little bit sloppy, but it gets you in the ballpark. The actual percentage is closer to 2.275%.

6 0

2 years ago

The formula for finding the perimeter of a rectangle is P = 2L + 2W. If a rectangle has a perimeter of 68 inches and the length

Artemon [7]

Answer:

Width = 10 inches

Step-by-step explanation:

Given the perimeter of a rectangle of 68 inches, and a length that is 14 inches longer than its width.

We can establish the following values to help us solve for the width of a rectangle:

Perimeter (P) = 68 inches

Length (L) = 14 + W inches

Width (W) = unknown

<h3 /><h3><u>Solve for the Width (W)</u></h3>

P = 2(L + W) ⇒ This is the same as P = 2L + 2W, except that 2 is factored out from the right-hand side.

Divide both sides by 2:

$\displaystyle\mathsf{\frac{P}{2}\:=\:\frac{2(L\:+\:W)}{2}}$

$\displaystyle\mathsf{\frac{P}{2}\:=L\:+\:W}$

Substitute the value of the Perimeter and the length (L) into the formula:

$\displaystyle\mathsf{\frac{68}{2}\:=14\:+W\:+\:W}$

Combine like terms on the right-hand side, and simplify the left-hand side of the equation:

$\displaystyle\mathsf{34\:=14\:+2W}$

Subtract 14 from both sides:

34 - 14 = 14 - 14 + 2W

20 = 2W

Divide both sides by 2 to solve for the width (W):

$\displaystyle\mathsf{\frac{20}{2}\:=\:\frac{2W}{2}}$

W = 10 inches

Therefore, the width of the rectangle is 10 inches.

<h3 /><h3><u>Double-check:</u></h3>

Verify whether the derived value for the width is correct:

P = 2L + 2W

68 = 2(14 + 10) + 2(10)

68 = 2(34) + 20

68 = 48 + 20

68 = 68 (True statement).

Thus, the length of the rectangle is 34 inches, and the width is 10 inches.

4 0

3 years ago

A park ranger has 32 feet of fencing to fence four sides of a rectangular recycling site. What is the greatest area of recycling

wariber [46]

The greatest area he can fence is 64 ft².

In order to maximize area and minimize perimeter, we use dimensions that are as close to equivalent as possible. 32 feet of fence for 4 sides gives us 8 feet of fence per side. We would have a square whose side length is 8; the area would be 8*8 = 64.

8 0

3 years ago

Read 2 more answers

The time needed to complete a final examination in a particular college course is normallydistributed with a mean of 80 minutes

Artist 52 [7]

Answer:

a) 0.023

b) 0.286

c) 10 students will be unable to complete the exam inthe allotted time.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 80 minutes

Standard Deviation, σ = 10 minutes

We are given that the distribution of time to complete an exam is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(completing the exam in one hour or less)

P(x < 60)

$P( x < 60) = P( z < \displaystyle\frac{60 - 80}{10}) = P(z < -2)$

Calculation the value from standard normal z table, we have,

$P(x < 60) =0.023= 2.3\%$

b) P(complete the exam in more than 60 minutes but less than 75 minutes)

$P(60 \leq x \leq 75) = P(\displaystyle\frac{60 - 80}{10} \leq z \leq \displaystyle\frac{75-80}{10}) = P(-2 \leq z \leq -0.5)\\\\= P(z \leq -0.5) - P(z < -2)\\= 0.309- 0.023 = 0.286= 28.6\%$

c) P(completing the exam in more than 90 minutes)

P(x > 90)

$P( x > 90) = P( z > \displaystyle\frac{90 -80}{10}) = P(z > 1)$

$= 1 - P(z \leq 1)$

Calculation the value from standard normal z table, we have,

$P(x > 90) = 1 - 0.8413 = 0.1587 = 15.87\%$

15.87% of children of class will require more than 90 minutes to complete the test.

Number of children =

$\dfrac{15.87}{100}\times 60 = 9.52\approx 10$

Approximately, 10 students of class will require more than 90 minutes to complete the test.

4 0

3 years ago