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balandron [24]
2 years ago
8

Solve for x. x/3 ≤ −6

Mathematics
2 answers:
ozzi2 years ago
5 0

x / 3 ≤ -6

---Multiply both sides by 3 to cancel out the denominator on the left side

x ≤ -18

Hope this helps!

omeli [17]2 years ago
3 0

Answer:

<h3>x=18 we multiply both 6and3</h3>
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timurjin [86]

Answer:

252 inches squared is the answer

Step-by-step explanation:

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3 years ago
Helppp with this question plzz
Gekata [30.6K]

Answer:

306

Step-by-step explanation:

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6 0
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!

5 0
3 years ago
Can someone please help?
Sati [7]

Answer:

It is closer to 7,000 than 8,000

4 0
3 years ago
Read 2 more answers
If the ordinate is three more than twice the abscissa and x E {-1,0,-1/3} which of the following sets of ordered pairs satisfies
almond37 [142]
<span>The ordinate is three more than twice the abscissa
</span>Ordnate: y; Abscissa: x
y=3+2x

x=-1→y=3+2(-1)=3-2→y=1→(x,y)=(-1,1)
x=0→y=3+2(0)=3+0→y=3→(x,y)=(0,3)
x=-1/3→y=3+2(-1/3)=3-2/3=(3*3-2)/3=(9-2)/3→y=7/3→(x,y)=(-1/3,7/3)

Answer: Option <span>A) {(-1,1),(0,3),(-1/3,7/3)}</span>
8 0
3 years ago
Read 2 more answers
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