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3 years ago

Liam has 3 games and 10 songs on his computer. How many outcomes are possible when randomly choosing a song and a game?

1 answer:

5 0

3 outcomes are possible because you only have 3 games on it, and it is mandatory to only have 1 game and 1 song at a time.

You might be interested in

If the slope of a line is 2 and its y-intercept is 5, what is the equation of the line?

CaHeK987 [17]

Hello,

y=2x+5
==========
(y-5)=2(x-0)

7 0

3 years ago

Read 2 more answers

Shirley wants to buy a skateboard for $64. She has $98 in her account. She spent $10.85 to buy stationary. She also wants to buy

klemol [59]

she had 98 but spent 10.85 so she now has:

98-10.85 = 87.15

skateboard cost 64

87.15 -64 = 23.15 is what she can spend on cookies

23.15 / 1.65 = 14.03 she can't buy more than 14 cookies

so answer is c. n ≤ 14

5 0

4 years ago

Read 2 more answers

The simple interest on a loan of $6000 for 3 years was $900. What was the rate of interest per annum

Thepotemich [5.8K]

Answer:

Step-by-step explanation:

900/3=300

300/6000=0.05

0.05=5%

7 0

3 years ago

In a contest to guess the number of jelly beans in a jar, Alice guessed 25;Betty, 18;Clare,23;Dotty,19;Edith,24;and Fran,17. Two

vampirchik [111]

Answer:

There are 21 jelly beans in the jar.

Step-by-step explanation:

Suppose, the number of jelly beans in the jar is $x$

Two were off by 2, two were off by 3, two were off by 4. That means, the six guessed number will be: $(x+2), (x-2), (x+3), (x-3), (x+4), (x-4)$

Given that, Alice guessed 25; Betty, 18; Clare, 23; Dotty, 19; Edith, 24; and Fran, 17.

So, the sum of all six guessed number will be: $25+18+23+19+24+17=126$

Thus, the equation will be..........

$(x+2)+(x-2)+(x+3)+(x-3)+(x+4)+(x-4)= 126\\ \\ 6x= 126\\ \\ x=\frac{126}{6}=21$

So, the number of jelly beans in the jar is 21.

5 0

3 years ago

CALCULUS: Determine which function is a solution to the differential equation y ' − y = 0.

Montano1993 [528]

C: none of these are solutions to the given equation.

• If y(x) = e², then y is constant and y' = 0. Then y' - y = -e² ≠ 0.

• If y(x) = x, then y' = 1, but y' - y = 1 - x ≠ 0.

The actual solution is easy to find, since this equation is separable.

y' - y = 0

dy/dx = y

dy/y = dx

∫ dy/y = ∫ dx

ln|y| = x + C

y = exp(x + C )

y = C exp(x) = C eˣ

8 0

3 years ago