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Len [333]
3 years ago
15

F(x)=lx+4l+3 how do u do it

Mathematics
1 answer:
dybincka [34]3 years ago
4 0

Answer:

implifying

f(x) = lx + 4l + -3

Multiply f * x

fx = lx + 4l + -3

Reorder the terms:

fx = -3 + 4l + lx

Solving

fx = -3 + 4l + lx

Solving for variable 'f'.

Move all terms containing f to the left, all other terms to the right.

Divide each side by 'x'.

f = -3x-1 + 4lx-1 + l

Simplifying

f = -3x-1 + 4lx-1 + l

Reorder the terms:

f = l + 4lx-1 + -3x-1

Step-by-step explanation:

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Answer:

C. Perpendicular Lines

D. Intersecting Lines

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Perpendicular lines are a point where two city roads intersect.

Hope this helps. I am sorry if you get it wrong.

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Which of the fallowing are proportions?
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Last years average ticket price was £22 this year the average ticket price has increased by 5% what is the average price now
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Exhibit 9-2 The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took the cu
Diano4ka-milaya [45]

Answer:

At a .05 level of significance, it can be concluded that the mean of the population is significantly more than 3 minutes.

Step-by-step explanation:

We want to test to determine whether or not the mean waiting time of all customers is significantly more than 3 minutes.

At the null hypothesis, we test if the mean is of at most 3 minutes, that is:

H_0: \mu \leq 3

At the alternative hypothesis, we test if the mean is of more than 3 minutes, that is:

H_1: \mu > 3

The test statistic is:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, \sigma is the standard deviation and n is the size of the sample.

3 is tested at the null hypothesis:

This means that \mu = 3

The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took the customers in the sample to check out was 3.1 minutes. The population standard deviation is known to be 0.5 minute.

This means that n = 100, X = 3.1, \sigma = 0.5

Value of the test statistic:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

z = \frac{3.1 - 3}{\frac{0.5}{\sqrt{100}}}

z = 2

P-value of the test and decision:

The p-value of the test is the probability of finding a sample mean above 3.1, which is 1 subtracted by the p-value of z = 2.

Looking at the z-table, z = 2 has a p-value of 0.9772.

1 - 0.9772 = 0.0228

The p-value of the test is of 0.0228 < 0.05, meaning that the is significant evidence to conclude that the mean of the population is significantly more than 3 minutes.

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