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MrRissso [65]
4 years ago
7

What is the explanation for 3 accidents in 12 months

Mathematics
1 answer:
Sever21 [200]4 years ago
8 0
This mean you need to be more careful in driving a car
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9+2×42+6-2<br><br>Please show the work​
timofeeve [1]

Answer:

The answer is 97

Step-by-step explanation:

According to PEMDAS, we will solve the multiplication first. 2 × 42 is 84. Next, we will add or subtract from left to right. 9 + 84 + 6 – 2 is 97, the answer to our problem.

8 0
2 years ago
Read 2 more answers
15 is what percent of 20
Lera25 [3.4K]
100%/x%=20/15
(100/X)*x=20/15*x   --I multiply both sides of the equation by X
100=1.33333333333*x --I divided both sides of the equation by                     .........................................(1.33333333333) to get X
100/1.33333333333=x
75=x
x=75

15 is 75% of 20

6 0
3 years ago
Select all angle measures for which cos0= 1/2
mrs_skeptik [129]

Answer:

cos -60° = \frac{1}{2} ,

cos 660° =  \frac{1}{2}

Step-by-step explanation:

hope it helps

6 0
3 years ago
3 less than m is greater than 9
Artist 52 [7]

Answer:

m-9 = 3 is the required expression

5 0
3 years ago
Read 2 more answers
The table below shows all outcomes for rolling 2 dice. The bold numbers 1 through 6 show the possible results for each die. All
Leokris [45]

Part I:

1. If you have to circle each outcome in the table that at least 1 die is a 4, then you have to circle row corresponding to the number 4 and column corresponding to the number 4.

2. If you have to cross out each outcome where the sum of the dice is greater than 5, then you have to cross out all not bold numbers 6, 7, 8, 9, 10, 11 and 12 in the table.

Part II:

Here you can see 36 possible outcomes.

1. The probability that at least 1 die is a 4 is:

Pr(A)=\dfrac{11}{36} - (favorable outcomes are 4+1=5, 4+2=6, 4+3=7, 4+4=8, 4+5=9, 4+6=10, 1+4=5, 2+4=6, 3+4=7, 5+4=9, 6+4=10).

2. The probability that the sum of the dice is greater than 5 is:

Pr(B)=\dfrac{26}{36}=\dfrac{13}{18}.

3. The probability that at least 1 die is a 4 and that the sum of the dice is greater than 5 is

Pr(A\cap B)=\dfrac{9}{36}=\dfrac{1}{4}.

4. The probability that at least 1 die is a 4 or that the sum of the dice is greater than 5 is:

Pr(A\cup B)=\dfrac{28}{36}=\dfrac{7}{9}.

4 0
4 years ago
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