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zloy xaker [14]
3 years ago
11

How do you create a model for 0.33 divided by 3

Mathematics
2 answers:
Triss [41]3 years ago
4 0
Since your chart is 100 squares you would want to shade 11 squares. Because your decimal is 11 out of 100.
Vlada [557]3 years ago
4 0
Shade 11 squares because that is 0.33/3 on the squares
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Anyone can help? Will give brainlist. I need this done before tomorrow :(​
Nitella [24]

Answer:

A - 17.576

Step-by-step explanation:

All we have to do is replace s with 2.6!

V = s^3

v = 2.6^2

v = (2.6)(2.6)(2.6)

v = 6.76 * 2.6

v = 17.576

So, the volume of the cube is 17.576 feet! :)

6 0
2 years ago
Experimental/theoretical
Lera25 [3.4K]

Answer:

A.200

Step-by-step explanation:

8 0
3 years ago
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3 groups of how many tenths is 1.5
topjm [15]
Well its 1 and 5 tenths 
- 1.5 = 15 tenths
- 5 groups of 3 is the answer
because 5 x 3 = 15
5 0
3 years ago
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According to Masterfoods, the company that manufactures M&M's, 12% of peanut M&M's are brown, 15% are yellow, 12% are re
Luda [366]

Answer:

The questions asked are

If you randomly select 4 peanuts

1. Compute the probability that exactly three of the four M&M’s are brown

2. Compute the probability that two or three of the four M&M’s are brown.

3. Compute the probability that at most three of the four M&M’s are brown.

4. Compute the probability that at least three of the four M&M’s are brown.

Step-by-step explanation:

Given the following information

Brown=12%. P(B)=0.12

Yellow=15%. P(Y)=0.15

Red=12%. P(R), =0.12

Blue=23%. P(B) =0.23

Orange, =23%. P(O) =0.23

Green=15%. P(G)=0.15

Question 1.

They are independent events

If there are exactly three brown and the last is not brown

P(B n B n B n B')

P(B)×P(B)×P(B)×P(B')

0.12×0.12×0.12×(1-P(B))

0.001728×(1-0.12)

0.001728×0.88

0.00152.

0.152%

2. If two or three are brown

I.e we are going to two brown and two none brown or three brown and one not brown. (P(B)×P(B)×P(B')×P(B'))+ (P(B)×P(B)×P(B'))

(0.12×0.12×0.88×0.88)+(0.12×0.12×0.12×0.88)

0.0112+0.00152

0.0127

1.27%

3. At most three brown out of four then we are going to have

BBBB', BBB'B', BB'B'B', B'B'B'B'

These are the cases of at most three brown.

P(B)×P(B)×P(B)×P(B') + P(B)×P(B)×P(B')×P(B') + P(B)×P(B')×P(B')×PB')+ P(B')×P(B')×P(B')×P(B')=

0.12×0.12×0.12×0.88+ 0.12×0.12×0.88×0.88+ 0.12×0.88×0.88×0.88+ 0.88×0.88×0.88×0.88=0.694

0.694

69.4%

4. At least 3 brown out of four selection

I.e BBBB', BBBB

These are the two options

P(B)×P(B)×P(B)×P(B') + P(B)×P(B)×P(B)×P(B)=

0.12×0.12×0.12×0.88 + 0.12×0.12×0.12×0.12

0.001728

0.173%

4 0
2 years ago
Read 2 more answers
Which of the following inequalities are correct?
Schach [20]

Answer:

Only C is correct

Step-by-step explanation:

59 is bigger than -59 not smaller so A is incorrect

0 is smaller than 76 not bigger do B is also incorrect

-76 is smaller than -59 according to number line so option C is correct

4 0
2 years ago
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