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Let f(x) = x^3 + 2x^2 + 3x + 1 and g(x)= 4x-5 Find f(x) x g(x)

Brums [2.3K]

$▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ { \huge \mathfrak{Answer}}▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪ ▪$

Given :

$f(x) = {x}^{3} + 2 {x}^{2} + 3x + 1$

$g(x) = 4x - 5$

let's solve for f(x) × g(x) :

$( {x}^{3} + 2x {}^{2} + 3x + 1)(4x - 5)$

$4x {}^{4} - 5 {x}^{3} + 8 {x}^{3 } - 10 {x}^{2} + 12 {x}^{2} - 15x + 4x - 5$

$4 {x}^{4} + 3 {x}^{3} + 2 {x}^{2} - 11x - 5$

therefore, correct choice is : D

8 0

2 years ago

If 2x + 5y = -5 what is x if y = -5 HELP ASAP

mash [69]

Answer:

x=10

Step-by-step explanation:

first substitute y with the value (-5)

2x+5(-5)=-5 which is simplified to 2x+-25=-5

add 25 to -5 which gives you 2x=20

then divide 20 by 2 which makes x=10

7 0

2 years ago

A salesperson works 40 hours per week at a job where she has two options for being paid. Option A is an hourly wage of 23$. Opt

Gwar [14]

Step-by-step explanation:

A salesperson works 40 hours per week at a job where she has two options for being paid. Option A is an hourly wage of 23$. Option B is a commission rate of 4% on weekly sales. How much does she need to sell this week to earn the same amount with the two options?
A salesperson works 40 hours per week at a job where she has two options for being paid. Option A is an hourly wage of 23$. Option B is a commission rate of 4% on weekly sales. How much does she need to sell this week to earn the same amount with the two options?
A salesperson works 40 hours per week at a job where she has two options for being paid. Option A is an hourly wage of 23$. Option B is a commission rate of 4% on weekly sales. How much does she need to sell this week to earn the same amount with the two options?

7 0

2 years ago

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Diane got a prepaid debit card with $25 on it. For her first purchase with the card, she bought some bulk ribbon at a craft stor

lorasvet [3.4K]

Answer:

48

Step-by-step explanation:

you can do 25 minus 18.76 then you get 6.24 then divide that by .13 and you get 48

3 0

2 years ago

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In order to ensure efficient usage of a server, it is necessary to estimate the mean number

juin [17]

Answer:

a. [36.19;39.21]

b. Reject the null hypothesis. The population mean of users that are connected at the same time is greater than 35.

Step-by-step explanation:

Hello!

Your study variable is,

X: "number of users of one server at a time"

The objective is to estimate the mean, for this, a sample of n=100 times was taken and the standard deviation S= 9.2 and the sample mean is X[bar]= 37.7 were calculated.

You need to study the population mean, for this you need your variable to have at least normal distribution. Since you don't have information about its distribution, but the sample is big enough (n≥30) you can apply the Central Limit Theorem and approximate the distribution of the sample mean X[bar] to normal:

X[bar]≈N(μ;σ²/n)

a. With this approximation, you can construct the 90% Confidence Interval using the approximate Z

[X[bar] ± $Z_{1-\alpha /2}$ * S/√n]

$Z_{1-\alpha /2}$ = $Z_{0.95}$ = 1.64

[37.7± 1.64* 9.2/√100]

[36.19;39.21]

b. You need to test if the population mean is greater than 35 with a level of significance of 1%.

The hypothesis is:

H₀: μ ≤ 35

H₁: μ > 35

α: 0.01

This is a one-tailed test so you have only one critical level (right tail):

$Z_{1\alpha } = Z_{0.99} = 2.33$

This means that if the value of the calculated statistic is equal or greater than 2.33 you will reject the null Hypothesis.

If the value is less than 2.33 you will support the null hypothesis.

The statistic is:

Z= X[bar] - μ = 37.7 - 35 = 2.93

S/√n 9.2/10

The value 2.93 > 2.33, so you reject the null hypothesis. This means that the population mean of users that are connected at the same time is greater than 35.

Note: To make the decision using the interval calculated on a), the hypothesis should have been two-tailed and the confidence and significance levels complementary.

I hope it helps!

7 0

2 years ago