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butalik [34]
3 years ago
15

24÷4=6 can be thought of as 24 broken into 4 groups of 6 each.

Mathematics
1 answer:
Marat540 [252]3 years ago
8 0

Answer:

yes

Step-by-step explanation:

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Solve the system of equations 2r + 2s = 50 and 2r – s = 17. A. r = 3, s = 11 B. r = 25, s = 33 C. r = 14, s = 11 D. r = –8, s =
scZoUnD [109]

2r = 17 + s = 50 - 2s \\ 3s = 50 - 17 = 33 \\ s = 11 \\  \\2 r = 50 - 2s \\ 2r = 50 - 2 \times 11 = 50 - 22 = 28 \\ r = 14




C. r=14, s =11
5 0
3 years ago
Paul has $78 in the bank and saves $30 each week. What is the initial value? Be sure to use negative signs if needed.
marin [14]

Answer:

The initial value is $78

Step-by-step explanation:

Given

Base\ Amount = \$78

Additional = \$30 (weekly)

Required

Determine the initial value

The initial value is the amount he has in its bank account before making his weekly savings.

From the question, we have that his initial balance is $78.

Hence, the initial value is $78

However, his weekly balance can be expressed as:

Balance = Base\ Amount + Additional * number\ of weeks

Represent number of weeks with x; So, we have:

Balance = 78 + 30 * x

Balance = 78 + 30 x

6 0
3 years ago
What is 2+2?!?!?!?!?!
Valentin [98]

Answer:

4

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
i have 17 coins. N of them are nickels and the rest are dimes. write an expression in two different ways for the amount of money
valkas [14]

Answer:

Step-by-step explanation:

(N)0.05 + (17-N)0.1 = M; M = amount of money I have.

Or 1.7-0.05N = M.

7 0
3 years ago
Backgammon is a board game for two players in which the playing pieces are moved according to the roll of two dice. players win
Ivahew [28]

The rolls of the dice are independent, i.e. the outcome of the second die doesn't depend in any way on the outcome of the first die.

In cases like this, the probability of two events happening one after the other is the multiplication of the probabilities of the two events.

So, the probability of rolling two 6s is the multiplication of the probabilities of rolling a six with the first die, and another six with the second:

P(\text{rolling two 6s}) = P(\text{rolling a 6}) \cdot P(\text{rolling a 6}) = \dfrac{1}{6} \cdot \dfrac{1}{6}  = \dfrac{1}{36}

Similarly,

P(\text{rolling two 3s}) = P(\text{rolling a 3}) \cdot P(\text{rolling a 3}) = \dfrac{1}{6} \cdot \dfrac{1}{6}  = \dfrac{1}{36}

Actually, you can see that the probability of rolling any ordered couple is always 1/36, since the probability of rolling any number on both dice is 1/6:

P(\text{rolling any ordered couple}) = P(\text{rolling the first number}) \cdot P(\text{rolling the second number}) = \dfrac{1}{6} \cdot \dfrac{1}{6}  = \dfrac{1}{36}

7 0
3 years ago
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