Synthetic division. (X^3+5x^2-x-9)+{x+2

Two communications companies offer calling plans. With Company X, it costs 30¢ to connect and then

tensa zangetsu [6.8K]

<h3>Answer: n+15</h3>

=======================================================

Explanation:

n = number of minutes
cost of company X = 3n+30
cost of company Y = 2n+15

To find out how much more company X charges, we subtract the two cost expressions

CompanyX - CompanyY = (3n+30)-(2n+15) = 3n+30-2n-15 = n+15 which is the final answer.

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An example:

Let's say you talk on the phone for n = 20 minutes.

Company X would charge you 3n+30 = 3*20+30 = 90 cents

Company Y would charge you 2n+15 = 2*20+15 = 55 cents

The difference of which is 90-55 = 35 cents.

If you plugged n = 20 into the n+15 expression we got, then n+15 = 20+15 = 35 matches up with the previous 35 cents.

This example helps confirm the answer. I'll let you try out other examples.

8 0

2 years ago

What is a common factor for the two fractional terms StartFraction 5 over 8 EndFraction x andStartFraction 11 over 8 EndFraction

Dafna1 [17]

Answer:

$\frac{1}{8x}$

Step-by-step explanation:

Given two fractional terms $\frac{5}{8x}\ and\ \frac{11}{8xy}$ . Their common factor is a value or function that can go in both fractional terms. The terms can be written as shown.

$\frac{5}{8x} = 5 *\frac{1}{8x}$

$\frac{11}{8xy} = 11 * \frac{1}{8x} * \frac{1}{y}$

It can be seen from the both equations that they both have $\frac{1}{8x}$ as one of their factors i.e <em>1/8x is common to both fractional terms</em>. This gives the common factor for the two fractional terms as $\frac{1}{8x}$

4 0

3 years ago

Read 2 more answers

A production process is checked periodically by a quality control inspector. the inspector selects simple random samples of 30 f

777dan777 [17]

Answer:

Population Mean = 2.0

Population Standard deviation = 0.03

Step-by-step explanation:

We are given that the inspector selects simple random samples of 30 finished products and computes the sample mean product weight.

Also, test results over a long period of time show that 5% of the values are over 2.1 pounds and 5% are under 1.9 pounds.

Now, mean of the population is given the average of two extreme boundaries because mean lies exactly in the middle of the distribution.

So, Mean, $\mu$ = $\frac{1.9+2.1}{2}$ = 2.0

Therefore, mean for the population of products produced with this process is 2.

Since, we are given that 5% of the values are under 1.9 pounds so we will calculate the z score value corresponding to a probability of 5% i.e.

z = -1.6449 {from z % table}

We know that z formula is given by ;

$Z = \frac{Xbar - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

-1.6449 = $\frac{1.9 - 2.0}{\frac{\sigma}{\sqrt{n} } }$ ⇒ $\frac{\sigma}{\sqrt{n} } = \frac{-0.1}{-1.6449}$

⇒ $\sigma =$ 0.0608 * $\sqrt{30}$ {as sample size is given 30}

⇒ $\sigma$ = 0.03 .

Therefore, Standard deviation for the population of products produced with this process is 0.0333.

7 0

3 years ago

10 Jenny has 12 marbles.

liraira [26]

Answer:

620

Step-by-step explanation:

12-1=11

11x50

1x70

add the two totals and you should get 620

Hope this helps!

5 0

3 years ago

Pumping stations deliver oil at the rate modeled by the function D, given by d of t equals the quotient of 5 times t and the qua

goblinko [34]

<h2>Hello!</h2>

The answer is: There is a total of 5.797 gallons pumped during the given period.

To solve this equation, we need to integrate the function at the given period (from t=0 to t=4)

The given function is:

$D(t)=\frac{5t}{1+3t}$

So, the integral will be:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dx$

So, integrating we have:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dt=5\int\limits^4_0 {\frac{t}{1+3t}} \ dx$

Performing a change of variable, we have:

$1+t=u\\du=1+3t=3dt\\x=\frac{u-1}{3}$

Then, substituting, we have:

$\frac{5}{3}*\frac{1}{3}\int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du\\\\\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u}{u} -\frac{1}{u } \ du$

$\frac{5}{9} \int\limits^4_0 {(\frac{u}{u} -\frac{1}{u } )\ du=\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u } )$

$\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u })\ du=\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du$

$\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du=\frac{5}{9} (u-lnu)/[0,4]$

Reverting the change of variable, we have:

$\frac{5}{9} (u-lnu)/[0,4]=\frac{5}{9}((1+3t)-ln(1+3t))/[0,4]$

Then, evaluating we have:

$\frac{5}{9}((1+3t)-ln(1+3t))[0,4]=(\frac{5}{9}((1+3(4)-ln(1+3(4)))-(\frac{5}{9}((1+3(0)-ln(1+3(0)))=\frac{5}{9}(10.435)-\frac{5}{9}(1)=5.797$

So, there is a total of 5.797 gallons pumped during the given period.

Have a nice day!

4 0

3 years ago