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Oxana [17]
4 years ago
9

Hello I need immediate help please?

Mathematics
1 answer:
weqwewe [10]4 years ago
4 0

Answer:

25

Step-by-step explanation:

;\

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PLEASE HELP ME OUT I NEED THIS ASAP, IM USING ALL THE POINTS I HAVE FOR THIS QUESTION
kogti [31]

Answer:

no you are not you are using 5 points!

Step-by-step explanation:

5 0
3 years ago
Here is a list of numbers 12 11 19 16 32 15 13​
ivann1987 [24]

Answer: You listed different numbers then in the picture since there is no 11  so this is based off of the picture.

Range: 20

Median: 15

Step-by-step explanation:

Range is just the highest number minus the lowest: 32-12 is 20

Median is the middle number after you put the numbers in order

12, 13, 13, 15, 16, 19, 32

3 0
2 years ago
In three days, Karen ate 1/3 of a large bag of chips during
Alexandra [31]

Answer:

Karen asked for the manager

Step-by-step explanation:

Karen asked for the maneger because that is what karens do. you’re lucky that you are not black, becuase she would falsely accuse you of murder after slightly bumping into her on accident.

6 0
3 years ago
Find the vertex for each equation:
tamaranim1 [39]

1. (−3,−1)

2. (3,−5)

3. (−3,−4)

5 0
3 years ago
An ordinary fair die is a cube with the numbers one through six on the sides represented by painted that. Imagine that such a di
Solnce55 [7]

Answer:

ok so what i think your trying to ask is if we roll two dice that the sum will be more then 6

Two dice

Assuming that the dice are unbiased or not " loaded".

Each side has the same probability, is 1/6 =0.16667, to turn up when rolled, if the die (D) is unbiased. The probability of a side turning up on D1 when 2 dice ( D1,D2) are rolled, is independent of the side turning up in D2. So this is an independent event.

How many ways can one get a sum total of 6 if D1 &D2 are rolled at the same time?

These are the possibilities

Case 1.

D1 =1 & D2=5

Or

D1= 5 & D2=1

Case 2.

D1 =2 & D2=4

Or

D1= 4 & D2= 2

Case 3. D1=3, D2=3

P3 =0.027778

Let's say, P 1 the probability for case 1 and P2 for case 2. There are no other cases.

The final probability P and is the sum total P = P1 + P2 + P3 the probability law of mutually exclusive events.

P1= 0.02778+ 0.02778 =0.055558

P2= 0.02778+0.02778 =0.055558

Same way,

P3=0.027778, when there is only one way to get the sum 6.

So, P = 0.138894

Based on truncating at the sixth decimal place.

A visual representation with two unbiased dice and the possible cases would also give the same result and is a short cut method. I like to derive from the basics.

Hope This Helps!!!

6 0
3 years ago
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