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Sphinxa [80]
3 years ago
7

Evaluate 1/5×5/6×3/7×21×16​

Mathematics
2 answers:
Alenkasestr [34]3 years ago
7 0

Answer:

24

Step-by-step explanation:

olga_2 [115]3 years ago
3 0

Answer:

24

Step-by-step explanation:

⅕×⅚×3/7×21×16

⅙×9×16 (Cancelling 5 by 5 and 21 by 3)

⅙×144

24

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Mr. Jones needs at least 600 paper plates for the school picnic. There are 75 plates in
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Answer:

The answer is 8 packages

Step-by-step explanation:

6 0
3 years ago
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Find the area of the shaded region. Express your answer in terms of π.
rusak2 [61]

Answer:

335.115

Step-by-step explanation:

First you find the area of the circles to do that you do pi times radius squared

For the top square do 3.14(pi)time 3²=28.26

For the middle square do 3.14 times 4.5²=63.585

For the bottom square do 3.14 times 6²=113.04

Now find the area of the entire thing do 18×12+9+6+3=18×30=540

Finally do 540-the area of the squares 28.26-63.585-113.04= 335.115

So the shaded region is 335.115

Hope this helps and have a great day!

6 0
3 years ago
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Two distinct number cubes, one red and one blue, are rolled together. Each number cube has sides numbered 1 through 6.
cestrela7 [59]

Answer:

P(E_1 or E_2) = \frac{7}{12}

Step-by-step explanation:

Given

Two cubes of side 1 - 6

Required

Probability that the outcome of the roll is an odd sum or a sum that is a multiple of 5

First, the sample space needs to be listed;

Let C_r represent the Red cube

C_b represent the Blue cube

S represent the sample space

C_r = (1,2,3,4,5,6)\\C_b = (1,2,3,4,5,6)\\S = (2,3,4,5,6,7,3,4,5,6,7,8,4,5,6,7,8,9,5,6,7,8,9,10,6,7,8,9,10,11,7,8,9,10,11,12)S is gotten by getting the sum of C_r and C_b

n(S) = 36

<em>Calculating the Probability</em>

Let E_1 represent the event that an outcome is an odd sum

E_1 = (3,5,7,3,5,7,5,7,9,5,7,9,7,9,11,7,9,11)

n(E_1) = 18

P(E_1) = \frac{n(E_1)}{n(S)}

P(E_1) = \frac{18}{36}

Let E_2 represent the event that an outcome is a multiple of 5

E_2 = (5,5,5,5,10,10,10)

n(E_2) = 7

P(E_2) = \frac{n(E_2)}{n(S)}

P(E_2) = \frac{7}{36}

Let E_3 represent the event that an outcome is an odd sum and a multiple of 5

E_3 = E_1 and E_2

E_3 = (5,5,5,5)

n(E_3) = 4

P(E_3) = \frac{n(E_3)}{n(S)}

P(E_3) = \frac{4}{36}

Calculating P(E_1 or E_2)

P(E_1 or E_2) = P(E_1) + P(E_2) - P(E_1 and E_2)

P(E_1 or E_2) = P(E_1) + P(E_2) - P(E_3)

P(E_1 or E_2) = \frac{18}{36} + \frac{7}{36} - \frac{4}{36}

P(E_1 or E_2) = \frac{18 + 7 - 4}{36}

P(E_1 or E_2) = \frac{21}{36}

P(E_1 or E_2) = \frac{7}{12}

Hence, the probability that the outcome of the roll is an odd sum or a sum that is a multiple of 5 is \frac{7}{12}

6 0
3 years ago
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kvv77 [185]
9.80 / 20 = 0.49.  Justin payed $0.49 per pound.  :P

3 0
3 years ago
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Tom and Sam sit down to eat a pie. Tom eats 2/3 of a pie and same eats 1/4 of the pie. What fraction of the pie is left?
Korolek [52]
The whole pie is 1, because 1 is 100%. Tom ate 2/3 of 1 and Sam ate 1/4 of the pie.
So, here's how to determine how much is left:
1 - (2/3 + 1/4) = 1 - (8/12 + 3/12) = 1 - 11/12 = 12/12 - 11/12 = 1/12

The correct answer is that 1/2 of the pie is left.
3 0
3 years ago
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