Answer:
The chance that there will be exactly 2 heads among the first five tosses and exactly 4 heads among the last 5 tosses is P=0.0488.
Step-by-step explanation:
To solve this problem we divide the tossing in two: the first 5 tosses and the last 5 tosses.
Both heads and tails have an individual probability p=0.5.
Then, both group of five tosses have the same binomial distribution: n=5, p=0.5.
The probability that k heads are in the sample is:
Then, the probability that exactly 2 heads are among the first five tosses can be calculated as:
For the last five tosses, the probability that are exactly 4 heads is:
Then, the probability that there will be exactly 2 heads among the first five tosses and exactly 4 heads among the last 5 tosses can be calculated multypling the probabilities of these two independent events:
Answer: SOH CAH TOA Tan C= 11/22= 1/2 So Tan C= 1/2
Sin B= x/17
Step-by-step explanation:
SOH CAH TOA = S=o/h C=a/h T=o/a.
I love trig ratios!
Hope this helped.
Answer:
They are at the same height at 1.13 seconds.
Step-by-step explanation:
Remark
The rockets are at the same height when f(x) = g(x) [see below] are the same. So you can equate them.
Givens
f(x) = - 16x^2 + 74x + 9
g(x) = -16x^2 + 82x I have changed this so you don't have 2 f(x)s
Solution
- f(x) = g(x)
- -16x^2 + 74x + 9 = -16x^2 + 82x Add: 16x^2 to both sides
- -16x^2+16x^2+74x + 9 = -16x^2+16x^2 + 82x Combine terms
- 74x + 9 = 82x Subtract 74x from both sides
- 74x - 74x + 9 = 82x - 74x Combine
- 9 = 8x Divide by 8
- 9/8 = 8x/8
- x = 1 1/8 Convert to decimal
- x = 1.125
- x = 1.13 [rounded]
Answer:
105
Step-by-step explanation:
(.5)πr²
to get the area of the shaded parts you have to subtract the area of the semicircle from the area of the rectangle
subtract the areas
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