All categories

2 years ago

This is the call plan for Robert's cell phone:

Mathematics

2 answers:

Oduvanchick [21]2 years ago

5 0

$0.20 times 30 minutes = $6.00
So $6.00 plus the $15 charge each month = $21.00

butalik [34]2 years ago

4 0

To answer this, first make an equation representing the data:

15+ 0.20X= Bill

now plug in data you know:

15 + 0.20(30)= bill

and finally solve it:
Bill= $21.00

You might be interested in

An object was dropped off the top of a building. The function

Lady_Fox [76]

What is the equation, it’s hard to understand

7 0

2 years ago

Read 2 more answers

A line passes through (5,3) with a slope of 3/5. Plot 3 points on this line

Aneli [31]

Answer:

y=3/5*x, A(0,0), B(1,3/5), C(2,6/5)

Step-by-step explanation:

A(5,3)=(x1,y1)....x1=5, y1=3

m=3/5

y-y1=m(x-x1)

y-3=3/5(x-5)

y-3=3/5*x-3/5*5

y-3=3/5*x - 3

y=3/5*x-3+3

y=3/5*x

x=0, y=3/5*x, y=3/5*0=0 A(0,0)

x=1, y=3/5*1=3/5 B(1,3/5)

x=2, y=3/5*2=6/5 C(2,6/5)

3 0

2 years ago

What is y=1/2x-10 Help me

Sidana [21]

Slope intercept form? Because slope intercept form is y= mx + b

7 0

2 years ago

Read 2 more answers

Can someone please help!!

Alchen [17]

Answer:

AB = 3 and AC = 4

Step-by-step explanation:

Given that the triangles are similar then the ratios of corresponding sides are equal, that is

$\frac{AB}{WX}$ = $\frac{BC}{WY}$ , substitute values

$\frac{AB}{30}$ = $\frac{3}{30}$ ( cross- multiply )

30AB = 90 ( divide both sides by 30 )

AB = 3

and

$\frac{AC}{XY}$ = $\frac{BC}{WY}$ , substitute values

$\frac{AC}{40}$ = $\frac{3}{30}$ ( cross- multiply )

30AC = 120 ( divide both sides by 30 )

AC = 4

3 0

2 years ago

Read 2 more answers

You roll a die with the sample space S = {1, 2, 3, 4, 5, 6}. You define A as {1 ,4, 6}, B as {1, 3, 4, 5, 6}, C as {1, 5}, and D

Lilit [14]

Answer:

Step-by-step explanation:

Recall that two events A,B are called mutually exclusive if and only if $A\cap B = \emptyset$ (their intersection is empty). They are exhaustive if they are mutually exclusive and their union is the sample space.

Based on this

a) Note that $A\cap B = \{1,6\}$ , so they are not mutually exclusive nor exhaustive.

b) $A\cap C = \{1\}$ so they are not mutually exclusive nor exhaustive.

c) $A\cap D = \emptyset$ , so they are mutually exclusive. Note that $A\cup D = \{1,2,3,4,5,6\}=S$ . Then they are exhaustive.

d) $B\cap C = \{1,5\}$ , so they are not mutually exclusive nor exhaustive.

6 0

3 years ago