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2 years ago

G(n) = 4n+5 f(n)= n^3+3n^2 find (g+f)(n)

Mathematics

1 answer:

Tpy6a [65]2 years ago

7 0

Given:

The two functions are:

$g(n)=4n+5$

$f(n)=n^3+3n^2$

To find:

The function $(f+g)(n)$ .

Solution:

We have,

$g(n)=4n+5$

$f(n)=n^3+3n^2$

Now,

$(f+g)(n)=f(n)+g(n)$

$(f+g)(n)=n^3+3n^2+4n+5$

Therefore, the correct option is B.

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25 is 45% of what number

Oksi-84 [34.3K]

Answer: 55.56

Step-by-step explanation:

Let the unknown number be k

(45 x k) / 100 = 25

45k/100 = 25

multiply both sides by 100

45k = 25 x 100

45k = 2500

divide both sides by 45

k = 2500/45

k = 55.56

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3 years ago

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Answer:

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Step-by-step explanation:

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2 years ago

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2 years ago

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kupik [55]

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Step-by-step explanation:

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2 years ago

Read 2 more answers

The number of "destination weddings" has skyrocketed in recent years. For example, many couples are opting to have their wedding

melamori03 [73]

Answer:

We conclude that the mean wedding cost is less than $30,000 as advertised.

Step-by-step explanation:

We are given the following data set:(in thousands)

29100, 28500, 28800, 29400, 29800, 29800, 30100, 30600

Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{236100}{8} = 29512.5$

Sum of squares of differences = 3408750

$S.D = \sqrt{\frac{3408750}{7}} = 697.82$

Population mean, μ = $30,000

Sample mean, $\bar{x}$ = $29512.5

Sample size, n = 8

Alpha, α = 0.05

Sample standard deviation, s = $ 697.82

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 30000\text{ dollars}\\H_A: \mu < 30000\text{ dollars}$ We use one-tailed t test to perform this hypothesis.

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{29512.5 - 30000}{\frac{697.82}{\sqrt{8}} } = -1.975$

Now,

$t_{critical} \text{ at 0.05 level of significance, 7 degree of freedom } = -1.894$

Since,

$t_{stat} < t_{critical}$

We fail to accept the null hypothesis and reject it.

We conclude that the mean wedding cost is less than $30,000 as advertised.

8 0

3 years ago