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zhuklara [117]
3 years ago
11

You are choosing between two different cell phone plans. The first plan charges a rate of 23 cents per minute. The second plan c

harges a monthly fee of $49.95 plus 9 cents per minute. How many minutes would you have to use in a month in order for the second plan to be preferable?
Mathematics
1 answer:
jenyasd209 [6]3 years ago
3 0

Answer:

357 minutes

Step-by-step explanation:

I subtracted 9 cents/minute from the 23 cents/minute to get 14 cents to get the difference between the two per minute charges.  I then divided the monthly cost of $49.95 by .14 to get 356.79...  So if you used 357 minutes in a month, the second plan would be 3 cents cheaper at $82.08 (.09 x 357= 32.13 + 49.95), vs. the first plan costing $82.11 (.23 x 357).  At 356 minutes the first plan would still be cheaper.

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!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Vesnalui [34]
Since they are independent events to find the probability of both is P(A) * P(B)

P(A) = P(Heads) = \frac{1}{2}
P(B) = P(Roll ≥ 4) = \frac{3}{6} =  \frac{1}{2}

Now multiply those fractions together

\frac{1}{2} *  \frac{1}{2}  =  \frac{1}{4} = P(Heads & ≥ 4)
5 0
3 years ago
In a small town, the use of coal (W in tons/per
zimovet [89]

Answer:

i)W = 2500 / T

ii)  W = 500 Tons

iii) grad W(10°) = - 25î

iv) The formulation is not practical

Step-by-step explanation:

i)  Write an equation describing the use of coal

As use of coal is inversely proportional to the average monthly temperature

if  W is use of coal in tons/per month then

W(t)  =   k / T              where k is a constant of proportionality and T is the average temperature in degrees. We have to determine k from given conditions

k  =  ??   we know that when   T  =  25°    W = 100 tons  the by subtitution

W = k/T           100  =  k /25       k  =  2500 Tons*degree

Then final equation is:

W = 2500 / T

ii) Find the amount of coal when T = 5 degrees

W = 2500 / 5

W = 500 Tons

iii)

The inverse proportionality implies that W will decrease as T increase.

The vector gradient of W function is:

grad W  = DW(t)/dt î  

grad W  = - 2500/T² î

Wich agrees with the fact that W is decreasing.

And when T = 100°

grad W(10°) =  - 2500/ 100 î      ⇒  grad W(10°) = - 25î

iv) When T =  0  The quantity of coal tends to infinite and the previous formulation is not practical

5 0
3 years ago
a parallelogram has vertices E(-3,1) F(0,2) G(-5,5) and H(-2,6) what are the coordinates of te midpoint of each diagonal
sesenic [268]

in parallelogram EFGH

the mid point of diagonal EG is

(-3-5)/2,(1+5)/2=(-4,3)

similarly the midpoint of diagonal FH is

(0-2)/2,(2+6)/2=(-1,4)

4 0
3 years ago
Find the mean of the following<br> data set.<br> 1,1,2,4,6,7,7,8,9,10,12,13,17,17,18
lisabon 2012 [21]

Answer:

8.8

Step-by-step explanation:

Add all numbers together to get 132

All number of numbers together to get 15

Divide the sum of all numbers by the number of numbers:

132/15

8.8

pls mark brainliest!

6 0
2 years ago
Data will be collected on the following variables. Which variable can be considered discrete? A. The height of a person B . The
kvv77 [185]

Answer: E: The number of books a person finished reading last month

Step-by-step explanation:

First, a discrete variable is a variable that only can take some given values in a set, the discrete variables are usually not dense, and a continuous variable is a variable that can take any value in a range (where the accepted values are dense).

So, for example, the set of the natural numbers is discrete, and the set of the real numbers can represent a continuous variable.

Here the only option that is really discrete will be the number of books that a person finished reading last month because here only positive whole numbers are accepted (you can not finish a 0.454 of a book)

The other options are continuous because all are classical measures.

For example, the weight of a person can jump between:

75.6kg and 75.7kg.

So you could think that this is discrete because the values between 75.6kg and 75.7g are not shown with our measuring device, but those will be added in the error of the measure because the weights between 75.6kg and 75.7kg are actually possible, so they must be accepted.

4 0
3 years ago
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