Louden County Wildlife Conservancy counts butterflies each year. Data over the last three years regarding four types ofbutterflies are shown below. What is the average number of Variegated Fritillaries for all three samples?
2009
Meadow Fritillary
Variegated Fritillary
Zebra Swallowtail
Eastern-Tailed Blue
Louden County Butterfly Count
2010
Meadow Fritillary
34
Variegated Fritillary 95
Zebra Swallowtail
Eastern-Tailed Blue 168
2011
Meadow Fritillary
Variegated Fritillary
Zebra Swallowtail
Eastern-Tailed Blue
14
63
10
170
|
33
242
55
165

Anwser choices
A)55
B)83
C)106
D)165

Answers

Answer 1

Answer:

Answer:

the answer is A

Step-by-step explanation:

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Megan buys 3 bracelets and 3 necklaces. Each bracelet costs 5. Megan pays on 40 and gets 4 change. what is the cost of one necklace?

Answers

The necklaces cost $3 each

What is the inverse for this problem?

Answers

replace f(x) with y:

y = 3x +2

interchange the variables:

x = 3y +2

switch the equation so the Y is on the left side:
3y +2 = x

subtract 2 from each side:
3y = x-2

divide each term by 3 to get y alone:

y = (x-2)/3

replace y with inverse of F(x) which is f^-1(x):

f^-1(x) = -(x-2)/3

Events A and B are mutually exclusive. Suppose event A occurs with probability 0.02 and event B occurs with probability 0.73. Compute the probability that B occurs or A does not occur (or both). Compute the probability that either B occurs without A occurring or A and B both occur

Answers

Answer:

1. The probability that B occurs or A does not occur (or both) is 0.73.

2. The probability that either B occurs without A occurring or A and B both occur is 0.73.

Step-by-step explanation:

It is given that the events A and B are mutually exclusive. It means the intersection of A and B is 0.

$P(A\cap B)=0$

Given information:

$P(A)=0.02$

$P(B)=0.73$

We get,

$P(A')=1-P(A)=1-0.02=0.98$

$P(B')=1-P(B)=1-0.73=0.27$

(1) We need to find the probability that B occurs or A does not occur (or both).

$P(B\cup A')+P(A\cap B)=P(B)+0$

$P(B\cup A')+P(A\cap B)=0.73+0$

$P(B\cup A')+P(A\cap B)=0.73$

Therefore the probability that B occurs or A does not occur (or both) is 0.73.

(2) We need to find the probability that either B occurs without A occurring or A and B both occur.

$P(B\cup A')+P(A\cap B)=P(B)+0$

$P(B\cup A')+P(A\cap B)=0.73+0$

$P(B\cup A')+P(A\cap B)=0.73$

Therefore the probability that either B occurs without A occurring or A and B both occur is 0.73.

Final answer:

For mutually exclusive events A and B, the probability that B occurs or A does not occur is approximately 1.45. The probability that either B occurs without A occurring or A and B both occur is 0.73 because A and B are mutually exclusive.

Explanation:

Events A and B are defined as mutually exclusive, which means they cannot occur at the same time. Hence, the probability that A and B both occur (referred to as P(A AND B)) is 0. In this question, for the first scenario, we need to compute the probability that B occurs or A does not occur which is denoted as P(B OR not A). Since events A and B are mutually exclusive, not A occurs with probability 1 - P(A) = 0.98. So, P(B OR not A) = P(B) + P(not A) - P(B AND not A) = 0.73 + 0.98 - (0.73 * 0.98) = 1.45 (approximately).

For the second scenario, we need to calculate the probability that either B occurs without A occurring or A and B both occur which is expressed as P((A and B) OR (B and not A)). But as we know P(A and B) = 0 for mutually exclusive events, there only remains P(B and not A). Again, as A and B are mutually exclusive, we can be sure that if B is happening, A is not, so the answer is P(B) = 0.73.