All categories

3 years ago

Jacky spent 53% of all her money to buy a computer game. How much did the game cost, if Jacky had $120 before buying the game?

Mathematics

2 answers:

Olegator [25]3 years ago

7 0

The game cost about $62.50.

Keith_Richards [23]3 years ago

5 0

The game cost 63.6$, because 53% of 120 is 63.6

You might be interested in

Type the solution.y=–3x–7y=x–3

Andrej [43]

Answer:

(x,y)=(-1,-4)

Step-by-step explanation:

y=–3x–7

y=x–3

(y=)-3x-7=x-3

-3x-x=-3+7

-4x=4

x=4/(-4)

x=-1

y=x-3

y=-1-3

y=-4

(x,y)=(-1,-4)

6 0

3 years ago

Need help due in 10 mins

Marina CMI [18]

Answer:

A) The amount of money deposited monthly

Step-by-step explanation:

y = 35x (slope) (amount deposited each month)+ 250 (y-intercept) (starting amount)

4 0

2 years ago

Which expression is equivalent to b^-2/ab^-5? a) a/b^5 b)1/ab^5 c)a^3b/1 d)b/a

Snezhnost [94]

$\bf a^{-{ n}} \implies \cfrac{1}{a^{ n}}\qquad \qquad
\cfrac{1}{a^{-n}}\implies a^{{ n}}\\\\
-------------------------------\\\\
\cfrac{b^{-2}}{ab^{-5}}\implies \cfrac{b^{-2}\cdot b^{+5}}{a}\implies \cfrac{b^{-2+5}}{a}\implies \cfrac{b^3}{a}$

4 0

3 years ago

Pls can someone help me. I m giving 10 pts . I may mark you as the braniliest

worty [1.4K]

100 notes were altogether

Solution:

Given that ratio of the number of $2 notes to the number of $5 notes was 4 : 1

number of $2 notes : number of $5 notes = 4 : 1

Let 4x be the number of $ 2 notes

Let 1x be the number of $ 5 notes

Given that total value of notes is $ 260

Therefore,

$ 2 (number of $ 2 notes ) + $ 5(number of $ 5 notes ) = $ 260

$ 2(4x) + $ 5(1x) = $ 260

8x + 5x = 260

13x = 260

x = 20

Thus number of notes altogether is given as:

4x + 1x = 4(20) + 1(20) = 80 + 20 = 100

Thus 100 notes were altogether

4 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

3 years ago