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3 years ago

Your cell phone company charges $20 per month for 500 minutes plus an additional

Mathematics

1 answer:

Alex_Xolod [135]3 years ago

8 0

Answer:

$21

Step-by-step explanation:

$20 for the 500 minutes

We still have to pay for 10 minutes.

10 cents per minute.

10centsx10 minutes.

100 cents, also known as 1 dollar.

$20 plus $1 means the answer is $21.

Hope this helps!

-AxekickNebulite

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QUESTION 3 Toyota provides an option of a sunroof and side airbag package for its Corolla model. This package costs $1400. Assum

Orlov [11]

Answer:

n=601

Step-by-step explanation:

Formula used:

$n=z(\frac{\alpha }{2})^2\frac{p(1-p) }{E^2}$

Solution:

$n=z(\frac{\alpha }{2})^2\frac{p(1-p) }{E^2}$

Where,

$\frac{\alpha }{2}=0.025$

As there is no previous estimate for p

Then, p=0.5

Here on using the table

$z(\frac{\alpha }{2}) =1.959963985$

Also,

E=0.04

p=0.5

Thus,

n=600.2279407

On approximating the value,

n=601

6 0

3 years ago

How much money will you have if you started with $1250 and put it in an account that earned 6.7% every year for 14 years?

KatRina [158]

Answer:

$3098.93

Step-by-step explanation:

We can use the formula for compound growth to solve this. The formula is:

$F=P(1+r)^t$

Where

F is the future value (the value at end of 14 years, our answer)

P is the initial amount invested ($1250)

r is the interest rate, in decimal (6.7% is 0.067)

t is the time in years (14, in our case)

<em>Plugging in all the information</em> we have:

$F=P(1+r)^t\\F=1250(1+0.067)^14\\F=1250(1.067)^14\\F=3098.93$

The account will accrue $3098.93 after 14 years.

3 0

3 years ago

(2pm^-1q^0)^-4 • 2m ^-1 p^3 / 2pq^2

Montano1993 [528]

Answer:

$\dfrac{m^3}{16p^2q^2}$

Step-by-step explanation:

Given:

$(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}$

$m^{-1}=\dfrac{1}{m}$

$q^0=1$

$2pm^{-1}q^0=2p\cdot \dfrac{1}{m}\cdot 1=\dfrac{2p}{m}$

$(2pm^{-1}q^0)^{-4}=\left(\dfrac{2p}{m}\right)^{-4}=\left(\dfrac{m}{2p}\right)^4=\dfrac{m^4}{(2p)^4}=\dfrac{m^4}{16p^4}$

$m^{-1}=\dfrac{1}{m}$

$2m^{-1} p^3=2\cdot \dfrac{1}{m}\cdot p^3=\dfrac{2p^3}{m}$

$\dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{\frac{2p^3}{m}}{2pq^2}=\dfrac{2p^3}{m}\cdot \dfrac{1}{2pq^2}=\dfrac{p^2}{mq^2}$

$(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{m^4}{16p^4}\cdot \dfrac{p^2}{mq^2}=\dfrac{m^3}{16p^2q^2}$

8 0

3 years ago

What is the square root of 74

laiz [17]

Answer:

8.602325

Simplified: 8.6

4 0

3 years ago

Find the minimum value if f(x) = xe^x over [-2,0]

SashulF [63]

Given function:

$f(x)=xe^x$

The minimum value of the function can be found by setting the first derivative of the function to zero.

$f^{\prime}(x)=xe^x+e^x$ $\begin{gathered} xe^x+e^x\text{ = 0} \\ e^x(x\text{ + 1) = 0} \end{gathered}$

Solving for x:

$\begin{gathered} x\text{ + 1 = 0} \\ x\text{ = -1} \end{gathered}$ $\begin{gathered} e^x\text{ = 0} \\ \text{Does not exist} \end{gathered}$

Substituting the value of x into the original function:

$\begin{gathered} f(x=1)=-1\times e^{^{-1}}_{} \\ =\text{ -}0.368 \end{gathered}$

Hence, the minimum value in the given range is (-1, -0.368)

6 0

1 year ago