Ask question

Ask question

All categories

3 years ago

7

Victor spend a total of 33 picture and text messages which cost 8.30.if text message rate is .18 per message and the picture mes

sage rate is .45 per picture how many of each type of message did he send

1 answer:

malfutka [58]3 years ago

7 0

Ok so

t=number of text messsages
p=number of picture messages

total number is 33
t+p=33

if text is 0.18 per pic and picture is 0.45 per text
total cost is 8.3
0.18t+0.45p=8.3
times 100 both sides for ease
18t+45p=830

we now have 2 equations
t+p=33
18t+45p=830

multiply first equation by -18 and add to second equation

18t+45p=830
<span>-18t-18p=-594 +</span>
0t+27p=236

27p=236
divide both sides by 27
p=8.740740740740744
this is impossible becasue you can't send 0.740740740740744 of a message so you did a mistype somewhere

You might be interested in

keri rented a canoe from a rental company for atotal of $35.00 the company charges $10.50 for the first hour and $8.25 per hour

Zielflug [23.3K]

Answer:

She rented it for 3 hours.

8.25 times 3 plus 10.50 = 35.00

6 0

2 years ago

Suppose that each child born is equally likely to be a boy or a girl. Consider a family with exactly three children. Let BBG ind

Gemiola [76]

Answer:

(a)

$S = \{GGG, GGB, GBG, GBB, BBG, BGB, BGG, BBB\}$

(b)

i.

$1\ girl = \{GBB, BBG, BGB\}$

$P(1\ girl) = 0.375$

ii.

$Atleast\ 2 \ girls = \{GGG, GGB, GBG, BGG\}$

$P(Atleast\ 2 \ girls) = 0.5$

iii.

$No\ girl = \{BBB\}$

$P(No\ girl) = 0.125$

Step-by-step explanation:

Given

$Children = 3$

$B = Boys$

$G = Girls$

Solving (a): List all possible elements using set-roster notation.

The possible elements are:

$S = \{GGG, GGB, GBG, GBB, BBG, BGB, BGG, BBB\}$

And the number of elements are:

$n(S) = 8$

Solving (bi) Exactly 1 girl

From the list of possible elements, we have:

$1\ girl = \{GBB, BBG, BGB\}$

And the number of the list is;

$n(1\ girl) = 3$

The probability is calculated as;

$P(1\ girl) = \frac{n(1\ girl)}{n(S)}$

$P(1\ girl) = \frac{3}{8}$

$P(1\ girl) = 0.375$

Solving (bi) At least 2 are girls

From the list of possible elements, we have:

$Atleast\ 2 \ girls = \{GGG, GGB, GBG, BGG\}$

And the number of the list is;

$n(Atleast\ 2 \ girls) = 4$

The probability is calculated as;

$P(Atleast\ 2 \ girls) = \frac{n(Atleast\ 2 \ girls)}{n(S)}$

$P(Atleast\ 2 \ girls) = \frac{4}{8}$

$P(Atleast\ 2 \ girls) = 0.5$

Solving (biii) No girl

From the list of possible elements, we have:

$No\ girl = \{BBB\}$

And the number of the list is;

$n(No\ girl) = 1$

The probability is calculated as;

$P(No\ girl) = \frac{n(No\ girl)}{n(S)}$

$P(No\ girl) = \frac{1}{8}$

$P(No\ girl) = 0.125$

7 0

3 years ago

Given the function f)(x) =-x^2+3x+8,determine the average rate of change of the function over the interval listed in the picture

Nadusha1986 [10]

Answer:

Vertex:

(

3

2

,

−

41

4

)

(

3

2

,

-

41

4

)

Focus:

(

3

2

,

−

10

)

(

3

2

,

-

10

)

Axis of Symmetry:

x

=

3

2

x

=

3

2

Directrix:

y

=

−

21

2

y

=

-

21

2

x

y

0

−

8

1

−

10

3

2

−

41

4

3

−

8

4

−

4

Vertex:

(

3

2

,

−

41

4

)

(

3

2

,

-

41

4

)

Focus:

(

3

2

,

−

10

)

(

3

2

,

-

10

)

Axis of Symmetry:

x

=

3

2

x

=

3

2

Directrix:

y

=

−

21

2

y

=

-

21

2

x

y

0

−

8

1

−

10

3

2

−

41

4

3

−

8

4

−

4

Step-by-step explanation:

7 0

3 years ago

Need help must bring back to my school August 8

ivolga24 [154]

2: 4 is a factor and multiple of 8. 8 and 4 is also even numbers so is is most likely to be divisible by any number

4 0

3 years ago

A car rental agency rents 190 cars per day at a rate of 29 dollars per day. for each 1 dollar increase in the daily rate, 5 fewe

Leviafan [203]

The equation that we can create from this situation is:

i = (190 – 5 x) * (29 + x)

where i is the income and x is the increase in daily rate

Expanding the equation:

i = 5510 + 190x – 145x - 5x^2

i = -5x^2 + 45x + 5510

Taking the 1st derivative:

di/dx = -10x +45

Set to zero to get the maxima:

-10x + 45 = 0

x = 4.5

So the cars should be rented at:

29 + x = 33.5 dollars per day

The maximum income is:

i = (190 – 5*4.5) * (33.5)

i = 5,611.25 dollars

7 0

3 years ago