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Marrrta [24]
3 years ago
5

Taxi A charges $0.20 per mile and an initial fee of $4. Taxi B charges $0.40 per mile and an initial fee of $2. Write an inequal

ity that can determine when the cost of Taxi B will be greater than Taxi A.
Mathematics
1 answer:
Paraphin [41]3 years ago
7 0
Well, for taxi A we have:

Cost of Taxi A = $ 0.20*X + $4
With "X" equal to "miles traveled".
For taxi B we have:
Cost of Taxi B = $ 0.40*X + $2
Then, when
Cost of Taxi B > Cost of Taxi A,

We Have:
$ 0.40*X + $2 > $ 0.20*X + $4

That is the inequality that you need!
Good luck!
M.
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3 years ago
Factor the trinomial in to 2 sets of parentheses <br> 12a squared minus 13a minus 35
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3 years ago
Which compound inequalities have the solution set represented by the given graph? Check all that apply
Contact [7]

We check with each options

'Or' represents the intersection of two graphs

'And' represents two separate graphs'

We have two separate shaded part in the given graph

So we ignore the options that has 'and' in between

LEts check first  and second option

\frac{x}{2}=13

Simplify the first part and second part

multiply both sides by 2 .

x < 2  or  4x - 2 > = 26

solve 4x-2 > = 26

add 2 on both sides and then divide both sides by 4

4x >= 28

x >= 7

So solution is x<2 or x>=7 . that is the graph on number line

Lets check with second option

3x-3<3  or  2x+8>=22

add 3 on both sides

3x < 6

divide both sides by 3

so x< 2

2x+8>=22

subtract 8 on both sides

2x >= 30

divide both sides by 2

x >= 15

x<2  or x>=15 that does not satisfies the graph

So option A is correct



7 0
3 years ago
Read 2 more answers
Show that the following functions are probability density functions for some value of k and determine k. Then, determine the mea
lord [1]

Answer:

a) 17.5

b) 15.6

c) 13.3

d) 21.51

Step-by-step explanation:

The given function is equal to:

f(x)=kx^2

where

\int\limits^y_0 {kx^{2} } \, =1

where y=23

Clearing k=0.00025

a) Ex=\int\limits^y_0 {xf(x)} \, dx =\int\limits^y_0 {x*0.00025x^{2} } \, dx =17.5

b)Vx=Ex^{2} -(Ex)^{2} =\int\limits^y_0 {x^{2}f(x) } \, dx-17.5^{2}  =\int\limits^y_0 {x^{2} *0.00025x^{2} } \, dx -17.5^{2} =321.82-306.25=15.6

c) The function is equal to:

f(x)=k(1+2x)

\int\limits^y_0 {k(1+2x)} \, =1

where y=20

k=0.0024

Ex=\int\limits^y_0 {xf(x)} \, dx =\int\limits^y_0 {x*0.0024(1+2x)} \, dx =13.3

d) Vx=Ex^{2} -(Ex)^{2} =\int\limits^y_0 {x^{2} f(x)} \, dx -13.3^{2}=\int\limits^y_0 {x^{2} *0.0024(1+2x)} \, dx-13.3^{2}   =198.4-176.89=21.51

8 0
3 years ago
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