Tan(35)=x/12
.7=x/12
12*.7=x
x=8.4
Answer: Our required probability is 0.3387.
Step-by-step explanation:
Since we have given that
Number of red cards = 4
Number of black cards = 5
Number of cards drawn = 5
We need to find the probability of getting exactly three black cards.
Probability of getting a black card = ![\dfrac{5}{9}](https://tex.z-dn.net/?f=%5Cdfrac%7B5%7D%7B9%7D)
Probability of getting a red card = ![\dfrac{4}{9}](https://tex.z-dn.net/?f=%5Cdfrac%7B4%7D%7B9%7D)
So, using "Binomial distribution", let X be the number of black cards:
![P(X=3)=^5C_3(\dfrac{5}{9})^3(\dfrac{4}{9})^2\\\\P(X=3)=0.3387](https://tex.z-dn.net/?f=P%28X%3D3%29%3D%5E5C_3%28%5Cdfrac%7B5%7D%7B9%7D%29%5E3%28%5Cdfrac%7B4%7D%7B9%7D%29%5E2%5C%5C%5C%5CP%28X%3D3%29%3D0.3387)
Hence, our required probability is 0.3387.
Answer:
x = 7, x = -3
Step-by-step explanation:
We can find the zeros by factoring the equation using the zero product property.
x^2-4x-21 = 0
(x-7)(x+3) = 0
<u>x-7 = 0</u>
x = 7
<u>x+3 = 0</u>
x = -3
Zeros are x=7 and x= -3.
Answer:
Step-by-step explanation:
a) The objective of the study is test the claim that the average gain in the green fees , lessons or equipment expenditure for participating golf facilities is less than $2,100 under the claim the null and alternative hypothesis are,
H₀ : μ = $2,100
H₀ : μ < $2,100
B) Suppose you selects α = 0.01
The probability that the null hypothesis is rejected when the average gain is $2,100 is 0.01
C) For α = 0.01
specify the rejection region of a large sample test
At the given level of significance 0.01 and the test is left-tailed then rejection level of a large-sample = < - 1.28
The answer is A. sin40=6/x