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-BARSIC- [3]
3 years ago
14

Michael pays his cell phone service provider $49.95 per month for 500 minutes. Any additional minutes used cost $0.15 each. In J

une, his phone bill is $61.20. How many additional minutes did he use?​
Mathematics
1 answer:
arsen [322]3 years ago
8 0

y = total cost used.

x = additional minutes

y = $49.95 + $0.15x

$61.20 = $49.95 + $0.15x. Subtract the $49.95 from each side.

$11.25 = $0.15x. Divide each side by $0.15

75 = x.

Michael used 75 additional minutes.

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One tenth the value of 2 in 8.524
Romashka [77]
85.25 would be one thenth of 8.524
3 0
3 years ago
Read 2 more answers
Here is a system of linear equations: y= x+4 y= -3x-8
ipn [44]

Answer:

  • A) (-3,1)

Step-by-step explanation:

<u>Given system of linear equations: </u>

  • y= x+4
  • y= -3x-8

---

  • x + 4 = -3x - 8
  • x + 3x = - 8 - 4
  • 4x = -12
  • x = -3

Then

  • y =-3 + 4 = 1

Solution is (-3, 1)

Correct option is A

5 0
3 years ago
B is the midpoint of AC. If Ab= x +5 and BC = 2x-11, find the measure of Ab.
LuckyWell [14K]

__________________________

Measurement of "AC" :

(x + 5) + (2x <span>− 11) ;
________________________

Find the measurement of "AB" [which is: "(x+5)" ]: 
______________________________________________
First, simplify to find the measurement of "AC" :
________________________________________
 </span>(x + 5) + (2x − 11) ;

=  (x + 5) + 1(2x − 11) ;

=   x + 5 + 2x − 11 ;

 → Combine the "like terms" ; 

   x + 2x = 3x ;
   
   5 − 11 = - 6 ;
______________
to get:  3x − 6 ; 
_______________
So,  (x + 5) + (2x − 11) =  3x − 6 ;
_______________________________
Solve for:  "(x + 5)"
_______________________________

We have: 
_______________________________

(x + 5) + (2x − 11) =  3x − 6  ;

Subtract:  "(2x − 11)" ;  from EACH SIDE of the equation ;
                                      to isolate "(x + 5)" on one side of the equation; 
                                      and to solve for "(x + 5)" ;
________________________________________________________
    →  (x + 5) + (2x − 11) − (2x − 11) =  (3x − 6) − (2x − 11) ;
  
         → (x + 5)  = (3x − 6) − (2x − 11) ;
_________________________________________________
  Note:  Simplify:  "(3x − 6) − (2x − 11)" ;

     →  (3x − 6) − (2x − 11)  ;
 
         =  (3x − 6) − 1(2x − 11) ;
 
         =   3x − 6 − 2x + 11 ;
__________________________
   →  Combine the "like terms" :
_____________________________
         +3x − 2x = 1x = x ;
  
          -6 + 11 = 5 ; 
_____________________________
To get:  x + 5 ; 

So we have:
______________________________
 x + 5 = x + 5 ; 
______________________________

So, x = all real numbers.

x = <span>ℝ </span>
4 0
3 years ago
The number of major earthquakes in a year is approximately normally distributed with a mean of 20.8 and a standard deviation of
SCORPION-xisa [38]

Answer:

a) 51.60% probability that in a given year there will be less than 21 earthquakes.

b) 49.35% probability that in a given year there will be between 18 and 23 earthquakes.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 20.8, \sigma = 4.5

a) Find the probability that in a given year there will be less than 21 earthquakes.

This is the pvalue of Z when X = 21. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{21 - 20.8}{4.5}

Z = 0.04

Z = 0.04 has a pvalue of 0.5160.

So there is a 51.60% probability that in a given year there will be less than 21 earthquakes.

b) Find the probability that in a given year there will be between 18 and 23 earthquakes.

This is the pvalue of Z when X = 23 subtracted by the pvalue of Z when X = 18. So:

X = 23

Z = \frac{X - \mu}{\sigma}

Z = \frac{23 - 20.8}{4.5}

Z = 0.71

Z = 0.71 has a pvalue of 0.7611

X = 18

Z = \frac{X - \mu}{\sigma}

Z = \frac{18 - 20.8}{4.5}

Z = -0.62

Z = -0.62 has a pvalue of 0.2676

So there is a 0.7611 - 0.2676 = 0.4935 = 49.35% probability that in a given year there will be between 18 and 23 earthquakes.

5 0
3 years ago
Lines p and q are parallel. what is the measure of angle 3 in degrees?
Dmitrij [34]
The correct answer is 27
8 0
3 years ago
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