All categories

3 years ago

Question 7 of 10

Mathematics

1 answer:

AURORKA [14]3 years ago

6 0

Answer:

b. there is not enough information to compare the mode

You might be interested in

Profit, P(x), is the difference between revenue, R(x), and cost, C(x), so P(x) = R(x) - CC). Which

aleksklad [387]

Answer:

Step-by-step explanation:

Profit, P(x), is the difference between revenue, R(x), and cost, C(x)

R(x) = 2x^4 - 3x^3 + 2x - 1

C(x) = x^4 -x^2 + 2x + 3

Substituting into the equation

P(x) = R(x) - C(x)

= (2x^4 - 3x^3 + 2x - 1) - (x^4 -x^2 + 2x + 3)

= x^4 - 3x^3 + x^2 - 4

The answer is the lower left option.

7 0

3 years ago

Read 2 more answers

HELP PLEASE:(

valentina_108 [34]

Answer:

i need more info

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Determine whether each set is finite or infinite. the set of fractions between 1 and 2.

yuradex [85]

The set of fractions between 1 and 2 are infinite.

3 0

4 years ago

What is The “Big Stick Diplomacy"

liberstina [14]

Answer:

"Big Stick diplomacy is the policy of carefully mediated negotiation ("speaking softly") supported by the unspoken threat of a powerful military ("big stick")."

Step-by-step explanation:

(nationalgeographic)

3 0

3 years ago

Read 2 more answers

Evaluate the sum of the following finite geometric series.

rjkz [21]

Answer:

$\large\boxed{\dfrac{156}{125}\approx1.2}$

Step-by-step explanation:

<h3>Method 1:</h3>

$\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}\\\\for\ n=1\\\\\left(\dfrac{1}{5}\right)^{1-1}=\left(\dfrac{1}{5}\right)^0=1\\\\for\ n=2\\\\\left(\dfrac{1}{5}\right)^{2-1}=\left(\dfrac{1}{5}\right)^1=\dfrac{1}{5}\\\\for\ n=3\\\\\left(\dfrac{1}{5}\right)^{3-1}=\left(\dfrac{1}{5}\right)^2=\dfrac{1}{25}\\\\for\ n=4\\\\\left(\dfrac{1}{5}\right)^{4-1}=\left(\dfrac{1}{5}\right)^3=\dfrac{1}{125}$

$\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}=1+\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{1}{125}=\dfrac{125}{125}+\dfrac{25}{125}+\dfrac{5}{125}+\dfrac{1}{125}=\dfrac{156}{125}$

<h3>Method 2:</h3>

$\sum\limits_{n=1}^4\left(\dfrac{1}{5}\right)^{n-1}\to a_n=\left(\dfrac{1}{5}\right)^{n-1}\\\\\text{The formula of a sum of terms of a geometric series:}\\\\S_n=a_1\cdot\dfrac{1-r^n}{1-r}\\\\r-\text{common ratio}\to r=\dfrac{a_{n+1}}{a_n}\\\\a_{n+1}=\left(\dfrac{1}{5}\right)^{n+1-1}=\left(\dfrac{1}{5}\right)^n\\\\r=\dfrac{\left(\frac{1}{5}\right)^n}{\left(\frac{1}{5}\right)^{n-1}}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\r=\left(\dfrac{1}{5}\right)^{n-(n-1)}=\left(\dfrac{1}{5}\right)^{n-n+1}=\left(\dfrac{1}{5}\right)^1=\dfrac{1}{5}$

$a_1=\left(\dfrac{1}{5}\right)^{1-1}=\left(\dfrac{1}{5}\right)^0=1$

$\text{Substitute}\ a_1=1,\ n=4,\ r=\dfrac{1}{5}:\\\\S_4=1\cdot\dfrac{1-\left(\frac{1}{5}\right)^4}{1-\frac{1}{5}}=\dfrac{1-\frac{1}{625}}{\frac{4}{5}}=\dfrac{624}{625}\cdot\dfrac{5}{4}=\dfrac{156}{125}$

5 0

3 years ago