This requires the Poisson distribution, where
area = 5-acres
and mean number of field mice = 12 (in 5-acres of field)
therefore
lambda=12 (mean, given)
and the probability of k mice in the 5-acre field is given by the Poisson distribution as
P(X=k)=lambda^k * e^(-lambda) / k! ..............(1)
To find the probability of having LESS than 7 field mice, we add the probabilities of 0 to 6, which is
P(X<7)=P(X=0)+P(X=1)+...+P(X=6)
evaluating with equation (1) for X=0 to 6, we get:
0 0.0000061 0.0000742 0.0004423 0.0017704 0.0053095 0.0127416 0.025481Total = 0.045822
Answer: The probability that fewer than 7 field mice are found in the 5-acre field is 0.0458.
Lets x = radius
x^2 + 10^2 = (x+8)^2
x^2 + 100 = x^2 + 16x + 64
16x = 100 - 64
16x = 36
x = 2.25
answer
B. 2.25
Answer:
184
Step-by-step explanation:
hope this helps!! :D
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Ok this inequality tells you the number of devices you can have before the new plan costs more than the old plan. The new plan expression is $4.50x + $94m = y ( total cost). The old plan is $175m = y (total cost). You can see m (number of months) in both equations, you don't need it this time since we're going to to compare both to one month. Since they're both equal to y you can make them equal to each other. $4.50x + $94 = $175. Now you want to figure when the new plan is less than the old plan you switch the equal sign for a less than sign. $4.50x + $94 < $175; this will help you find the inequality you want. From there just use algebraic steps to find that x has to less than 18 or
x < 18.
