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disa [49]
3 years ago
5

(a) the probability that a vampire bat would pick A (b) the probability that a vampire bat would pick C (c) the probability that

a vampire bat would pick T
Mathematics
1 answer:
grin007 [14]3 years ago
3 0

Answer:

See Explanation

Step-by-step explanation:

See attachment for complete question.

From the attachment, we have that:

A = 71

C = 121

T = 66

First, we calculate the total

Total = A + C + T

Total = 71 + 121 + 66

Total = 258

Solving (a): Probability of A

This is calculated using:

Probability = \frac{A}{Total}

Probability = \frac{71}{258}

Probability = 0.275

Solving (b): Probability of C

This is calculated using:

Probability = \frac{C}{Total}

Probability = \frac{121}{258}

Probability = 0.469

Solving (b): Probability of T

This is calculated using:

Probability = \frac{T}{Total}

Probability = \frac{66}{258}

Probability = 0.256

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Answer:

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3 0
3 years ago
How to solve X - 1v+ 4 + 7y - 3
Nostrana [21]
The first step to solving this is to remember our mathematical rules. They state that when the term has a coefficient of -1,, the number doesn't have to be written but the sign needs to remain. This will change the expression to the following:
x - v + 4 + 7y - 3
Now subtract the numbers 4 and 3 from each other.
x - v + 1 + 7y 
Since this expression cannot be simplified any further,, the correct answer to your question would be x - v + 1 + 7y.
Let me know if you have any further questions.
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A certain species of animal is endangered and it’s numbers are decreasing 12% annually. There are currently 230 animals in the p
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B:f(x)=230(0.88);121 animals


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3 years ago
Write a function G whose graph represents a reflection in the y-axis of the graph of f (x) = |2x–1|+3
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Step-by-step explanation:

8 0
3 years ago
Consider the equation below.
Korvikt [17]

Answer:

Equation in square form:

y=3(x+5)^2-4

Extreme value:

(h,k)=(-5,-4)

Step-by-step explanation:

We are given

y=3x^2+30x+71

we can complete square

y=3(x^2+10x)+71

we can use formula

a^2+2ab+b^2=(a+b)^2

y=3(x^2+2\times x\times 5)+71

now, we can add and subtract 5^2

y=3(x^2+2\times x\times 5+5^2-5^2)+71

y=3(x^2+2\times x\times 5+5^2)-3\times 5^2+71

y=3(x+5)^2-75+71

So, we get equation as

y=3(x+5)^2-4

Extreme values:

we know that this parabola

and vertex of parabola always at extreme values

so, we can compare it with

y=a(x-h)^2+k

where

vertex=(h,k)

now, we can compare and find h and k

y=3(x+5)^2-4

we get

h=-5

k=-4

so, extreme value of this equation is

(h,k)=(-5,-4)

6 0
4 years ago
Read 2 more answers
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