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3 years ago

A jet plane costs an airline $65,500,00. What is the place value of the digit 5 in this number?

1 answer:

5 0

|Ten Mils| Mils|100 thousands|10 thousands|Thousands|100's|Tens|Units |
|-----------|------|--------------------|------------------|---------------|-------|---------------|
| 6 | 5 | 5 | 0 | 0 | 0 | 0 |

You can find easily the solution

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Is x=4 a solution to the equation 9x = 32 + x?

madam [21]

Answer:

yes

Step-by-step explanation:

because 9*4=36 and well it's 36=32*? so 36-32 is 4

So well the solution is x=4

8 0

3 years ago

Read 2 more answers

4. Adrian and Rory started comic book collections at the same time. Adrian

PilotLPTM [1.2K]

As both are adding same number of comic books each months, they will not have same number of books. Adrian will have 4 more comic books.

Step-by-step explanation:

Given,

Adrian starts with 12 comic books and add 4 each month.

Let,

m be the number of books.

A(m) = 4m+12

Rory starts with 8 comic books and add 4 comic books each month.

R(m) = 4m+8

As both are adding same number of comic books each months, they will not have same number of books. Adrian will have 4 more comic books.

6 0

4 years ago

For $7.52, you purchased 8 pens and highlighters from a local bookstore. Each highlighter cost $1.09 and each pen cost $.69. How

seraphim [82]

i can. it says you got 8 pens. 1.09h + .69(8) = 7.52 1.09h + 5.52 = 7.52 do you know what to do from here?

5 0

3 years ago

in a standard casino dice game the roller wins on the first roll if he rolls a sum of 7 or 11. what is the probability of winnin

Alborosie

Answer-

The probability of winning on the first roll is 0.22

Solution-

As in the game of casino, two dice are rolled simultaneously.

So the sample space would be,

$|S|=6^2=36$

Let E be the event such that the sum of two numbers are 7, so

E = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}

$|E|=56$

$\therefore P(E)=\dfrac{|E|}{|S|}=\dfrac{6}{36}$

Let F be the event such that the sum of two numbers are 11, so

F = {(6,5), (5,6)}

$|F|=2$

$\therefore P(F)=\dfrac{|F|}{|S|}=\dfrac{2}{36}$

Now,

$P(\text{sum is 7 or 11)}=P(E\ \cup\ F)=P(E)+P(F)=\dfrac{6}{36}+\dfrac{2}{36}=\dfrac{8}{36}=\dfrac{2}{9}=0.22$

8 0

4 years ago

Create an equivalent system of these equations and test your solution x + y = 1 x 3y =9

ad-work [718]

An equvilent equation
remember you can do anything to an equation as long asyou do it to both sides

assuming yo have
x+y=1 and
x-3y=9
mulitply both by 2
2x+2y=2
2x-6y=18
those are equvilent

ok, solve initial

x+y=1
x-3y=9
multiply first equation by -1 and add to 2nd equation

-x-y=-1
x-3y=9 +
0x-4y=8

-4y=8
divide both sides by -4
y=-2

sub back
x+y=1
x-2=1
add 2
x=3

x=3
y=-2
(3,-2)

if we test it in other one

2x+2y=2
2(3)+2(-2)=2
6-4=2
2=2
yep

2x-6y=18
2(3)-6(-2)=18
6+12=18
18=18
yep

solution is (3,-2)

4 0

4 years ago

Read 2 more answers