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Rus_ich [418]
3 years ago
8

A farm lets you pick 3 pints of raspberries for $12 how much does it cost per pint

Mathematics
1 answer:
user100 [1]3 years ago
5 0

Answer:

$4

Step-by-step explanation:

1 pint = $4

$4 × 3 = $12

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Write an equation of a line wit the given slope and y intercept
Orlov [11]

Answer:

Step-by-step explanation:

the standard equation of a line is in the form y=mx+b where m is the slope and b is the y intercept so if m = 1/4 and b = -3/4 you can just replace those spots giving you y=1/4x+-3/4

6 0
3 years ago
Adrian's CD player can hold six disks at a time and shuffles all of the albums and their songs. If he has thirteen CD's, how man
Crank
Adrian can put a total of 1716 combinations in the player. 

The formula to be used is as follows:

total combinations = n!/[r! *(n-r)!]
n = number of disks available  = 13
r = number of disks that be held = 6

= 13! = (13• 12•11•10•9•8•7•6•5•4•3•2•1)/(6! <span>• 7!)
</span>=1716

Thank you for posting your question. Feel free to ask me more.



6 0
3 years ago
Simplify the expression<br><br> -10r+-4r
Anna [14]
The Answer to this solution is -14r
8 0
2 years ago
Evaluate |-3|<br><br> ±3<br> -3<br> 3
marin [14]

Answer:

Its just 3

Step-by-step explanation:

Because its absolute value.

Hope this helps!

4 0
3 years ago
Read 2 more answers
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
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