Can someone help me im in the test rn

Mason's car has a gas tank that holds 24 gallons of gas. The tank is exactly 1/4 full. How many gallons of gas are in Mason's ca

Helga [31]

Its going to be 6 gallons of gas

7 0

3 years ago

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Which linear function includes the points (3,9) and (-2,-11)

Murrr4er [49]

$General\ equation\ for\ line\ in\ slope\ intercept\ form:\\\\y=ax+b\\\\To\ find\ a\ and\ b\ substitude\ points\ (3,9),\ (-2,-11)\ into\ equation\\\\ \left \{ {{9=3a+b}\ \ \ \ \atop {-11=-2a+b\ |*-1}} \right.\\\\ \left \{ {{9=3a+b}\ \ \ \ \atop {11=2a-b}} \right.\\+----\\\\Addition\ method\\\\20=5a\ \ |:5\\\\a=4\\ b=9-3a\\b=9-3*4=9-12=-3\\\\Answer\ y=4x-3
$

7 0

4 years ago

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Vacation Property Rentals purchased 50 television sets for their rental units at a cost of $250 per television. The televisions

Alex787 [66]

Under A straight line basis which is a method of computing depreciation and amortization by dividing the difference between an asset's cost and its expected salvage value by the number of years it is expected to be used. Also known as straight line depreciation or straight line amortization, this is the simplest depreciation method. But instead of that find The rate of depreciation
100/5 years=20% depreciation rate per year
Total cost 250×50=12,500
Salvage value 40×50=2,000
Subtract the salvage value from the total cost of televisions
12,500−2,000=10,500
In the first year the depreciation is
10,500×0.2=2,100
Book value
12,500−2,100=10,400
In the second year the depreciation is
10,500×0.2=2,100
Book value
10,400−2,100=8,300
In the third year the depreciation is
10500×0.2=2100
Book value
8300-2100=6200

the book value for all of the televisions at the end of the third year is 6200

8 0

4 years ago

A common way for two people to settle a frivolous dispute is to play a game of rock-paper-scissors. In this game, each person si

Leokris [45]

Answer:

(a) P(A) = 0.34

(b) P(B) = 0.33

(c) P(C) = 0.33

(d) The Complement of event A = 1 - P(A) = 0.66

Step-by-step explanation:

We are given that in the long run, roommate A chooses rock 36% of the time, and roommate B chooses rock 22% of the time; roommate A selects paper 32% of the time, and roommate B selects paper 25% of the time; roommate A chooses scissors 32% of the time, and roommate B chooses scissors 53% of the time.

Let the probability that roommate A chooses rock = P( $R_A$ ) = 0.36

The probability that roommate A chooses paper = P( $P_A$ ) = 0.32

The probability that roommate A chooses scissors = P( $S_A$ ) = 0.32

The probability that roommate B chooses rock = P( $R_B$ ) = 0.22

The probability that roommate B chooses paper = P( $P_B$ ) = 0.25

The probability that roommate B chooses scissors = P( $S_B$ ) = 0.53

(a) Let A = event that roommate A wins the game and thus does not have to wash the dishes.

This will happen only when roommate A chooses rock and roommate B chooses scissors or roommate A chooses paper and roommate B chooses rock or roommate A chooses scissors and roommate B chooses paper.

So, P(A) = $P(R_A) \times P(S_B) + P(P_A) \times P(R_B) + P(S_A) \times P(P_B)$

= (0.36 $\times$ 0.53) + (0.32 $\times$ 0.22) + (0.32 $\times$ 0.25)

= 0.1908 + 0.0704 + 0.08

P(A) = 0.34

(b) Let B = event that roommate B wins the game and thus does not have to wash the dishes.

This will happen only when roommate B chooses rock and roommate A chooses scissors or roommate B chooses paper and roommate A chooses rock or roommate B chooses scissors and roommate A chooses paper.

So, P(B) = $P(R_B) \times P(S_A) + P(P_B) \times P(R_A) + P(S_B) \times P(P_A)$

= (0.22 $\times$ 0.32) + (0.25 $\times$ 0.36) + (0.53 $\times$ 0.32)

= 0.0704 + 0.09 + 0.1696

P(B) = 0.33

(c) Let C = event that the game ends in a tie.

This will happen only when roommate A chooses rock and roommate B also chooses rock or roommate A chooses paper and roommate B also chooses paper or roommate A chooses scissors and roommate B also chooses scissors.

So, P(C) = $P(R_A) \times P(R_B) + P(P_A) \times P(P_B) + P(S_A) \times P(S_B)$