Your answer is 89.56 hope this helps and God bless you
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:

I'd start by writing an equation for each of the right triangles. (Pythagorean theorem)
y² + 9² = z²
x² + z² = (4+9)²
4² + y² = x²
we want to find z so combine the equations by substituting the other variables x,y out.
substitute y² for (x² - 4²) in 1st equation.
(x² - 4²) + 9² = z²
now by rearranging the 2nd equation we can substitute x² for (13² - z²)
(13² - z²) - 4² + 9² = z²
169 - z² - 16 + 81 = z²
234 - z² = z²
234 = 2z²
234/2 = z²
117 = z²
√(117) = z
√(9*13) = z
3√(13) = z
13 goes in the box
Answer:
the third answer
Step-by-step explanation:
c = 4
c + 1 = 1 + 1 + 1 + 1 + 1
4 + 1 = 1 + 1 + 1 + 1 + 1
5 = 5
Answer:
1/-9
Step-by-step explanation:
hope it helps... ✌️✌️✌️✌️
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