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3 years ago

If a person sees 7 pennies, 6 Nickels, 4 dimes, and 6 Quarters, and he picks 1 of each, how many different combos can there be?

1 answer:

3 0

Well there really wouldn't be a combo since everything is different so I'm saying nun? that's what I'm going with. ( If this is wrong correct me )

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The mean temperature for the first 7 days in January was 3 °C.

weeeeeb [17]

$m = \frac{a_1 + a_2 + ... + a_n}{n}\\~\\m = \frac{3 * 7 + 2}{8} = \frac{21 + 2}{8} = \frac{23}{8} = 2,875$

Answer: 2,875.

6 0

3 years ago

line p contains points A(-7,-9) and B(4,0) line q is parallel to line p line r is perpendicular to line q what is the slope of l

igomit [66]

Use the slope formula (y2-y1)/(x2-x1)
(0+9)/(4+7)=9/11
Since q is parallel to r the slope will be the same. 9/11 is the slope of line r

4 0

3 years ago

ipn [44]

Answer:

50 soldiers must be transferred elsewhere.

Step-by-step explanation:

We solve this question by proportions, using a rule of three.

As the number of soldiers decrease, the provisions last for more time. This means that the measures are inversely proportional, and we have an inverse rule of three, using line multiplication, instead of cross.

30 days of provisions - 200 soldiers

40 days of provisions - x soldiers

$40x = 200*30$

Simplifying by 40

$x = 5*30 = 150$

The provisions will last for 40 days with 150 soldiers, which means that 200 - 150 = 50 soldiers must be transferred elsewhere.

4 0

3 years ago

Simplify this expression 4m^4n^4/14m

EastWind [94]

It would simplify down to (2m^3n^4)/7

5 0

3 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) = $P(R_A) \times P(R_B) + P(P_A) \times P(P_B) + P(S_A) \times P(S_B)$

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

= 0.0792 + 0.08 + 0.1696

P(C) = 0.3288 ≈ 0.33

(d) The complement of event A = P(A') = 1 - P(A)

= 1 - 0.34 = 0.66.

4 0

3 years ago