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9966 [12]
2 years ago
9

If p(a)=0.2, p(b)=0.2, and a and b are mutually exclusive, are they independent?

Mathematics
1 answer:
creativ13 [48]2 years ago
5 0

he events are not independent

<h3>How to determine if they are mutually exclusive?</h3>

The given parameters are:

P(A) = 0.2

P(B) = 0.2

The events are mutually exclusive.

So, as have

P(A or B) = P(A) + P(B)

If the events are independent, then

P(A and B) = P(A) + P(B) - P(A or B)

P(A and B) = P(A) * P(B)

So, we have

P(A and B) = 0.2 + 0.2 - 0.2 - 0.2 = 0

P(A and B) = 0.2 * 0.2 = 0.04

The above equations are not equal

Hence, the events are not independent

Read more about probability at:

brainly.com/question/24756209

#SPJ1

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Write the slope-intercept form of the equation for the line.
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Well, let's take a peek at the graph of that line, hmmm let's pick two points, heck, those tow on the extremes anyway, and those are (-5,2) and (5,-1), alrite.. now, let's do some checking on that.

\bf \begin{array}{lllll}&#10;&x_1&y_1&x_2&y_2\\&#10;%   (a,b)&#10;&({{ -5}}\quad ,&{{ 2}})\quad &#10;%   (c,d)&#10;&({{ 5}}\quad ,&{{ -1}})&#10;\end{array}&#10;\\\\\\&#10;% slope  = m&#10;slope = {{ m}}= \cfrac{rise}{run} \implies &#10;\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{-1-2}{5-(-5)}\implies \cfrac{-1-2}{5+5}&#10;\\\\\\&#10;\cfrac{-3}{10}

\bf \stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-2=-\cfrac{3}{10}[x-(-5)]&#10;\\\\\\&#10;y-2=-\cfrac{3}{10}(x+5)\implies y-2=-\cfrac{3}{10}x-\cfrac{3}{2}\implies y=-\cfrac{3}{10}x-\cfrac{3}{2}+2&#10;\\\\\\&#10;y=-\cfrac{3}{10}x+\cfrac{1}{2}
5 0
3 years ago
Please I need this ​
Free_Kalibri [48]

Answer:

1) not

2) QE

3) QE

4) QE

5) QE

6) QE

7) QE

8) not

9) QE

10) not

hope this helps you!

7 0
3 years ago
Population Growth A lake is stocked with 500 fish, and their population increases according to the logistic curve where t is mea
Alexus [3.1K]

Answer:

a) Figure attached

b) For this case we just need to see what is the value of the function when x tnd to infinity. As we can see in our original function if x goes to infinity out function tend to 1000 and thats our limiting size.

c) p'(t) =\frac{19000 e^{-\frac{t}{5}}}{5 (1+19e^{-\frac{t}{5}})^2}

And if we find the derivate when t=1 we got this:

p'(t=1) =\frac{38000 e^{-\frac{1}{5}}}{(1+19e^{-\frac{1}{5}})^2}=113.506 \approx 114

And if we replace t=10 we got:

p'(t=10) =\frac{38000 e^{-\frac{10}{5}}}{(1+19e^{-\frac{10}{5}})^2}=403.204 \approx 404

d) 0 = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}

And then:

0 = 7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)

0 =19e^{-\frac{t}{5}} -1

ln(\frac{1}{19}) = -\frac{t}{5}

t = -5 ln (\frac{1}{19}) =14.722

Step-by-step explanation:

Assuming this complete problem: "A lake is stocked with 500 fish, and the population increases according to the logistic curve p(t) = 10000 / 1 + 19e^-t/5 where t is measured in months. (a) Use a graphing utility to graph the function. (b) What is the limiting size of the fish population? (c) At what rates is the fish population changing at the end of 1 month and at the end of 10 months? (d) After how many months is the population increasing most rapidly?"

Solution to the problem

We have the following function

P(t)=\frac{10000}{1 +19e^{-\frac{t}{5}}}

(a) Use a graphing utility to graph the function.

If we use desmos we got the figure attached.

(b) What is the limiting size of the fish population?

For this case we just need to see what is the value of the function when x tnd to infinity. As we can see in our original function if x goes to infinity out function tend to 1000 and thats our limiting size.

(c) At what rates is the fish population changing at the end of 1 month and at the end of 10 months?

For this case we need to calculate the derivate of the function. And we need to use the derivate of a quotient and we got this:

p'(t) = \frac{0 - 10000 *(-\frac{19}{5}) e^{-\frac{t}{5}}}{(1+e^{-\frac{t}{5}})^2}

And if we simplify we got this:

p'(t) =\frac{19000 e^{-\frac{t}{5}}}{5 (1+19e^{-\frac{t}{5}})^2}

And if we simplify we got:

p'(t) =\frac{38000 e^{-\frac{t}{5}}}{(1+19e^{-\frac{t}{5}})^2}

And if we find the derivate when t=1 we got this:

p'(t=1) =\frac{38000 e^{-\frac{1}{5}}}{(1+19e^{-\frac{1}{5}})^2}=113.506 \approx 114

And if we replace t=10 we got:

p'(t=10) =\frac{38000 e^{-\frac{10}{5}}}{(1+19e^{-\frac{10}{5}})^2}=403.204 \approx 404

(d) After how many months is the population increasing most rapidly?

For this case we need to find the second derivate, set equal to 0 and then solve for t. The second derivate is given by:

p''(t) = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}

And if we set equal to 0 we got:

0 = \frac{7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)}{(1+19e^{-\frac{t}{5}})^3}

And then:

0 = 7600 e^{-\frac{t}{5}} (19e^{-\frac{t}{5}} -1)

0 =19e^{-\frac{t}{5}} -1

ln(\frac{1}{19}) = -\frac{t}{5}

t = -5 ln (\frac{1}{19}) =14.722

7 0
3 years ago
Solve<br> the quadratic equation<br> 4X^2+ 3x-27=0
maxonik [38]

Answer:

x = -3 or 9/4

Step-by-step explanation:

(x+3)(4x-9)

or use the quadratic formula

5 0
3 years ago
Traverion wants to find the answer to the question “How much money does the average college professor make?” He surveys 50 profe
Grace [21]

Answer: The results do not represent the population because the sample does not represent professors from different areas.


Step-by-step explanation:

Since Traverion wants to find the answer to the question “How much money does the average college professor make?”

He only surveys 50 professors from the university in his hometown.

Although he should survey professors from different areas but he only perform his survey only a specific university in his hometown.

Therefore, the survey is biased as the sample size is too small.

Hence, The results do not represent the population because the sample does not represent professors from different areas.

4 0
3 years ago
Read 2 more answers
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