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balandron [24]
2 years ago
5

Which graph is the result of reflecting f(x) = One-fourth(8)x across the y-axis and then across the x-axis? On a coordinate plan

e, an exponential function approaches y = 0 in quadrant 2 and incresaes into quadrant 1. It goes through the y-axis at (0, 0.25) and goes through (1, 2). On a coordinate plane, an exponential function approaches y = 0 in quadrant 1 and increases in quadrant 2. It crosses the y-axis at (0, 0.25) and goes through (negative 1, 2). On a coordinate plane, an exponential function approaches y = 0 in quadrant 3 and decreases into quadrant 4. It crosses the y-axis at (0, negative 0.25) and goes through (1, negative 2). On a coordinate plane, an exponential funtion increases in quadrant 3 into quadrant 4 and approaches y = 0. It goes through (negative 1, negative 2) and crosses the y-axis at (0, negative 0.25).
Mathematics
2 answers:
VladimirAG [237]2 years ago
8 0

Answer:

Its B

Explanation:

I just answered it.

scoundrel [369]2 years ago
7 0

Answer:

Which graph is the result of reflecting f(x) = One-fourth(8)x across the y-axis and then across the x-axis?

On a coordinate plane, an exponential function approaches y = 0 in quadrant 2 and incresaes into quadrant 1. It goes through the y-axis at (0, 0.25) and goes through (1, 2).

On a coordinate plane, an exponential function approaches y = 0 in quadrant 1 and increases in quadrant 2. It crosses the y-axis at (0, 0.25) and goes through (negative 1, 2).

On a coordinate plane, an exponential function approaches y = 0 in quadrant 3 and decreases into quadrant 4. It crosses the y-axis at (0, negative 0.25) and goes through (1, negative 2).

On a coordinate plane, an exponential funtion increases in quadrant 3 into quadrant 4 and approaches y = 0. It goes through (negative 1, negative 2) and crosses the y-axis at (0, negative 0.25).

Step-by-step explanation:

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Answer:

Step-by-step explanation:

the answer is 67

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Use the figure below to estimate the indicated derivatives, or state that they do not exist. If a derivative does not exist, ent
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F(x)=3x/2 for 0≤x≤2 
<span>.....=6 - 3x/2 for 2<x≤4 </span>
<span>g(x) = -x/4 + 1 for 0≤x≤4 and g'(x)=-1/4 </span>
<span>so h(x)= f(g(x)) = (3/2)(-¼x+1)=-3x/8 + 3/2 for 0≤x≤2 </span>
<span>for x=1, h'(x)=-3/8 so h'(1)=-3/8 </span>

<span>When x=2, g(2)=1/2 so h'(2)=g'(2)f '(1/2)= -(1/4)(3/2)=-3/8 </span>

<span>When x=3, h(x)=6 - (3/2)(1 - x/4) = 9/2 +3x/8 </span>
<span>h'(x)=3/8 so h'(3) = 3/8</span>
8 0
2 years ago
Read 2 more answers
Two percent of the customers of a store buy cigars. Half of the customers who buy cigars buy beer. 25 percent who buy beer buy c
enyata [817]

Answer:

0.95

Step-by-step explanation:

The computation of the probability that a customer neither buys beer nor buys cigars is given below;

Given that, the probabilities are

The customers who purchased cigars be 0.02

The customers who purchased cigars + beer 0.50

And, the customers who purchased beer + cigars be 0.25

Now the probabilities where the customer purchased both

= 0.05 × 0.02

= 0.10

The probability where the customer purchased beer is

= 0.01 ÷ 0.25

= 0.04

Now the probability where a customer neither buys beer nor buys cigars is

= 1 - 0.02 + 0.04 - 0.01

= 0.95

4 0
1 year ago
The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.3 minutes and a standard deviation of 3.3
In-s [12.5K]

Answer:

a) There is a 74.22% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

b) There is a 1-0.0548 = 0.9452 = 94.52% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

c) There is a 68.74% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.3 minutes and a standard deviation of 3.3 minutes. This means that \mu = 8.3, \sigma = 3.3.

(a) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes?

We are working with a sample mean of 37 jets. So we have that:

s = \frac{3.3}{\sqrt{37}} = 0.5425

Total time of 320 minutes for 37 jets, so

X = \frac{320}{37} = 8.65

This probability is the pvalue of Z when X = 8.65. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{8.65 - 8.3}{0.5425}

Z = 0.65

Z = 0.65 has a pvalue of 0.7422. This means that there is a 74.22% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

(b) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes?

Total time of 275 minutes for 37 jets, so

X = \frac{275}{37} = 7.43

This probability is subtracted by the pvalue of Z when X = 7.43

Z = \frac{X - \mu}{\sigma}

Z = \frac{7.43 - 8.3}{0.5425}

Z = -1.60

Z = -1.60 has a pvalue of 0.0548.

There is a 1-0.0548 = 0.9452 = 94.52% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

(c) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes?

Total time of 320 minutes for 37 jets, so

X = \frac{320}{37} = 8.65

Total time of 275 minutes for 37 jets, so

X = \frac{275}{37} = 7.43

This probability is the pvalue of Z when X = 8.65 subtracted by the pvalue of Z when X = 7.43.

So:

From a), we have that for X = 8.65, we have Z = 0.65, that has a pvalue of 0.7422.

From b), we have that for X = 7.43, we have Z = -1.60, that has a pvalue of 0.0548.

So there is a 0.7422 - 0.0548 = 0.6874 = 68.74% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes.

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