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Tju [1.3M]
3 years ago
12

/Algebra/ please help, I'll help you if I can.​

Mathematics
1 answer:
viva [34]3 years ago
6 0

Answer:

the answer is b and you can help by being my girl friend

Step-by-step explanation:

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Select the correct word form for 87,055​
sveticcg [70]

Answer: Eighty-seven thousand-fifty five

Step-by-step explanation:

3 0
2 years ago
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Find the value of x at which the function has a possible relative maximum or minimum point.​ (Recall that e Superscript x is pos
shutvik [7]

Answer:

x= -11/4 is a maximum.

Step-by-step explanation:

Remember that a function has its critical points where the derivative equal zero. Therefore we need to compute the derivative of this function and find the points where the derivative is zero. Using the chain rule and the product rule we get that

f'(x)=  -e^{-4x}(11+4x)

And then we get that   if   11+4x = 0   then   x = -11/4 . So it has a critical point at   x = -11/4.

Now, if the second derivative evaluated at that point is less than 0 then the point is a maximum and if is greater than zero the point is a minimum.

Since

f''(x) = 8e^{-4x} (5+2x)\\f''(-11/4) = -239496.56

x= -11/4 is a maximum.

6 0
3 years ago
*nervous laugh* SO ummm geometry isnt my thing actually im rather bad it please help!
Savatey [412]
I believe the answer would be D 92 and 88 hope it helped 
8 0
3 years ago
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The prior probabilities for events A1 and A2 are P(A1) = 0.35 and P(A2) = 0.50. It is also known that P(A1 ∩ A2) = 0. Suppose P(
forsale [732]

Answer:

Step-by-step explanation:

Hello!

Given the probabilities:

P(A₁)= 0.35

P(A₂)= 0.50

P(A₁∩A₂)= 0

P(BIA₁)= 0.20

P(BIA₂)= 0.05

a)

Two events are mutually exclusive when the occurrence of one of them prevents the occurrence of the other in one repetition of the trial, this means that both events cannot occur at the same time and therefore they'll intersection is void (and its probability zero)

Considering that P(A₁∩A₂)= 0, we can assume that both events are mutually exclusive.

b)

Considering that P(BIA)= \frac{P(AnB)}{P(A)} you can clear the intersection from the formula P(AnB)= P(B/A)*P(A) and apply it for the given events:

P(A_1nB)= P(B/A_1) * P(A_1)= 0.20*0.35= 0.07

P(A_2nB)= P(B/A_2)*P(A_2)= 0.05*0.50= 0.025

c)

The probability of "B" is marginal, to calculate it you have to add all intersections where it occurs:

P(B)= (A₁∩B) + P(A₂∩B)=  0.07 + 0.025= 0.095

d)

The Bayes' theorem states that:

P(Ai/B)= \frac{P(B/Ai)*P(A)}{P(B)}

Then:

P(A_1/B)= \frac{P(B/A_1)*P(A_1)}{P(B)}= \frac{0.20*0.35}{0.095}= 0.737 = 0.74

P(A_2/B)= \frac{P(B/A_2)*P(A_2)}{P(B)} = \frac{0.05*0.50}{0.095} = 0.26

I hope it helps!

5 0
3 years ago
Find the value of x. Then classify the triangle
vredina [299]
So basically what you have to do is
5 0
3 years ago
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