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madam [21]
3 years ago
8

Help me with this plz

Mathematics
1 answer:
Dmitrij [34]3 years ago
7 0

Answer:

the third one

Step-by-step explanation:

We can break it up into (-8*3)*(x*x^2). This means that it is -24*x^3, or -24x^3

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100 POINTS PLZ HELP
crimeas [40]

An ordered pair such as (3,-1), is a shorthand way of writing two variables, such as x = 3 and y = -1. The order of the numbers in the pair is important: x always comes before y.

An ordered pair is a composition of the x coordinate (abscissa) and the y coordinate (ordinate), having two values written in a fixed order within parentheses.

It helps to locate a point on the Cartesian plane for better visual comprehension.

The numeric values in an ordered pair can be integers or fractions.

Ordered Pair = (x,y)

Where, x = abscissa, the distance measure of a point from the primary axis “x”

And, y = ordinate, the distance measure of a point from the secondary axis “y”

In the Cartesian plane, we define a two-dimensional space with two perpendicular reference lines, namely x-axis and y-axis. The point where the two lines meet at “0” is the origin.

HOPE IT HELPS

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What is the solution to the inequality +4 50?
kicyunya [14]

Answer:

d<-1.75

Step-by-step explanation:

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3 years ago
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!

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Answer:

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