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babymother [125]
3 years ago
7

To win a game, roger needs to score 2000 points. So far he has scored 837 points how many more points does roger need to score?

Mathematics
1 answer:
BartSMP [9]3 years ago
5 0
Roger needs to score 1163 more points to win the game.


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Find the 31st term of the sequence 12,16,20
Natali [406]

Answer:

136

Step-by-step explanation

multiply 4 by 31 and add 20

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4 years ago
ASAP I need help,please
kompoz [17]

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C

Step-by-step explanation:

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Randy walks his dog each morning he walks 7/12 of a mile in 7 minutes how many miles does he walk in 1 minute
Andrej [43]

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Tim and Nikko bought new computers. Tim spent $55 more than 2/3 of the amount that Nikko spent for his computer. The total amoun
juin [17]

Answer:

<h2>C.  $1,689.00</h2>

Step-by-step explanation:

Let the amount spent by Tim be x and the amount spent by Nikko be y

If Tim spent $55 more than 2/3 of the amount that Nikko spent for his computer, then amount spent by Tim will be x = 55+2/3 y

Since the total amount that Tim and Nikko spent for their computers was $2,870 then;

x+y = $2,870

Substituting  x = 55+2/3y into the equation above, we will have;

55+2y/3 + y = 2,870

(165 + 2y)/3 + y = 2,870

(165 + 2y+3y)/3 = 2,870

cross multiply

165 + 2y+3y =  3*2870

165+5y = 8610

5y = 8610-165

5y = 8445

y = 8445/5

y = 1689

<em>Hence Nikko spent  a sum of $1,689.00 on his computer.</em>

4 0
4 years ago
Timothy creates a game in which the player rolls 4 dice. What is the probability in this game of having exactly two dice or more
devlian [24]

Answer:

B. 0.132

Step-by-step explanation:

For each time the dice is thrown, there are only two possible outcomes. Either it lands on a five, or it does not. The probability of a throw landing on a five is independent of other throws. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

Timothy creates a game in which the player rolls 4 dice.

This means that n = 4

The dice can land in 6 numbers, one of which is 5.

This means that p = \frac{1}{6}

What is the probability in this game of having exactly two dice or more land on a five?

P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4)

In which

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 2) = C_{4,2}.(\frac{1}{6})^{2}.(\frac{5}{6})^{2} = 0.116

P(X = 3) = C_{4,2}.(\frac{1}{6})^{3}.(\frac{5}{6})^{1} = 0.015

P(X = 4) = C_{4,4}.(\frac{1}{6})^{4}.(\frac{5}{6})^{0} = 0.001

P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4) = 0.116 + 0.015 + 0.001 = 0.132

So the correct answer is:

B. 0.132

8 0
3 years ago
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