Can you help me w/ #6 plz
1 answer:
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Answer:
since we have 6 witnesses and 5 men so,
therefore the probability that they would pick the same man is to 5 decimal places
Step-by-step explanation:
probability is the likelihood of an event occurring or not happening.
The easiest way to solve this is to divide the shape into 2 rectangles.
The first rectangle being the big one on the left with the 12 and 6
and the second one being the little one on the right, next to it with the 5
So 12 x 6 = 72 (dont pay attention to the 7, 5, or 10, because all we need is 1 length and 1 width)
And for the smaller one it is a little bit trickier.
So on the bottom we see that is says 10ft, but we cant use that because what we really need is the measurement between the 5 and 7 on the top of the little rectangle
At the top we see that it says 6ft
We can do 10-6 to find the missing measurement.
Because the length of the whole like is 10, and the length of one of the little lines is 6, so the length of 6 + some number =10
So 10-6 = 4
Which means the number we need is 4
So now we have 5 x 4 = 20
72 + 20 = 92
Answer: 92
I also attatched an image, I hope this helps! Ask me if you have any other questions about this problem!
The first rectangle being the big one on the left with the 12 and 6
and the second one being the little one on the right, next to it with the 5
So 12 x 6 = 72 (dont pay attention to the 7, 5, or 10, because all we need is 1 length and 1 width)
And for the smaller one it is a little bit trickier.
So on the bottom we see that is says 10ft, but we cant use that because what we really need is the measurement between the 5 and 7 on the top of the little rectangle
At the top we see that it says 6ft
We can do 10-6 to find the missing measurement.
Because the length of the whole like is 10, and the length of one of the little lines is 6, so the length of 6 + some number =10
So 10-6 = 4
Which means the number we need is 4
So now we have 5 x 4 = 20
72 + 20 = 92
Answer: 92
I also attatched an image, I hope this helps! Ask me if you have any other questions about this problem!