Answer:
the answer is A because 3 + 5 is 8 :) made a 100 btw :3
Step-by-step explanation:
There are 81 girls to every 72 boys in total, which can be represented by 81/72, which put into decimal form, is 1.125. If you then take the number of boys in the group, 16, and multiply it by this number (because the problem states that the ratio is constant) you can find the number of girls in the group.
16*1.125=18
So there are 18 girls in a group with 16 boys.
You can't really tell anything without the picture of the graph its self
A)The y-intercept is 50 because when X=0, y=50. This shows that the car was moving at 50 mph as an initial speed.
B) in two hours (from 1 to 3) the speed increased by 4 miles per 2hours. (from 52 to 56). 4/2 = 2 mph, which is the average rate of change. This represents the amount of speed increased each hour.
C) y=2x+50, and y=60 in this case, so 60=2x+50
60=2x+50
-50 -50
-----------------
10=2x
5=x, so the domain is X<=5 (less than or equal to 5)
The probability that it rains at most 2 days is 0.00005995233 and the variance is 0.516
<h3>The probability that it rains at most 2 days</h3>
The given parameters are:
- Number of days, n = 7
- Probability that it rains, p = 95%
- Number of days it rains, x = 2 (at most)
The probability that it rains at most 2 days is represented as:
P(x ≤ 2) = P(0) + P(1) + P(2)
Each probability is calculated as:

So, we have:



So, we have:
P(x ≤ 2) =0.00000002097 + 0.00000168821 + 0.00005824315
P(x ≤ 2) = 0.00005995233
Hence, the probability that it rains at most 2 days is 0.00005995233
<h3>The mean</h3>
This is calculated as:
Mean = np
So, we have:
Mean = 7 * 92%
Evaluate
Mean = 6.44
Hence, the mean is 6.44
<h3>The standard deviation</h3>
This is calculated as:
σ = √np(1 - p)
So, we have:
σ = √7 * 92%(1 - 92%)
Evaluate
σ = 0.718
Hence, the standard deviation is 0.718
<h3>The variance</h3>
We have:
σ = 0.718
Square both sides
σ² = 0.718²
Evaluate
σ² = 0.516
This represents the variance
Hence, the variance is 0.516
Read more about normal distribution at:
brainly.com/question/4079902
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