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Neko [114]
3 years ago
5

Perfect square numbers

Mathematics
1 answer:
Softa [21]3 years ago
4 0

Answer:

All the sides are equal. any number as long as all the sides have the same number.

Step-by-step explanation:

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Aubrey claims that if the dimensions of the parallelogram shown are doubled, then the area of the larger parallelogram will be 4
Ostrovityanka [42]

Answer:

Aubrey is correct because the area of the new parallelogram is 12 (6) = 72 square inches. The original area is 18 square inches. Since 4 (18) = 72, the new parallelogram has 4 times the area of the original.

Step-by-step explanation:

A parallelogram with a base of 6 inches and height of 3 inches. A side has a length of 5 inches.

Aubrey is correct because the area of the new parallelogram is 12 (6) = 72 square inches. The original area is 18 square inches. Since 4 (18) = 72, the new parallelogram has 4 times the area of the original.

Aubrey is correct because the area of the new parallelogram is 10 (7) = 70 square inches. The original area is 18 square inches. Since 4 (18) = 72, it is about 4 times larger than the original.

Aubrey is incorrect because if one doubles each dimension, then the area will automatically be doubled as well. The original area is 18 square inches so the new parallelogram will have an area of 2 (18) = 36, or two times more than the original.

Aubrey is incorrect because if one doubles each dimension, then the area will automatically be doubled as well. The original area is 30 square inches so the new parallelogram will have an area of 2 (30) = 60, or 2 times more than the original.

Workings

Area of a parallelogram=base×height

Original parallelogram

Base=6 inches

Height=3 inches

Area=base×height

=6×3

=18 square inches

New parallelogram with doubled dimensions

Base=6 inches doubled=12 inches

Height=3 inches doubled=6 inches

Area=base×height

=12×6

=72 square inches

New area=4 times original area

New area=4×18

New area=72 square inches

5 0
3 years ago
I need help ASAP please!!!
Aleksandr-060686 [28]

Answer: The answer is b 100+(-500)

Step-by-step explanation: If you solve for the original and this answer, they will both equal -400

8 0
3 years ago
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Solving exponential growth and decay problems
Paul [167]
See tha attached pdf file and let me know how useful this answer is.
Download pdf
6 0
3 years ago
In a casino game, gamblers are allowed to roll n fair, 6-sided dice. If a 6 shows up on any of them, the gambler gets nothing. I
Soloha48 [4]

Answer:

a) \bf \boxed{W(n) = 3*n*\left( \frac{5}{6} \right)^n\;dollars}

the smallest n that maximizes the expected payoff is n=5.

b) 4

Step-by-step explanation:

a)

The expected amount of $ won for each die would be the average of 1, 2, 3, 4 and 5 which is $3.

Let W(n) the expected money won when rolling n dice.  

n =1

If the gambler rolls only one die, the expected money won would be $3 times the probability of not getting a 6, which is 5/6.

So  

\bf W(1) = 3*1*\frac{5}{6}

n=2

If the gambler rolls 2 dice, the expected money won would be $3 times the probability of not getting a 6 in any of the dice. Since the outcome of the rolling does not depend on the previous rollings, the probability is  

\bf \frac{5}{6}\times\frac{5}{6}=\left( \frac{5}{6} \right)^2

and

\bf W(2) = 3*2*\left( \frac{5}{6} \right)^2

n=3

Similarly, since the probability of not getting a 6 in 3 dice equals

\bf \left( \frac{5}{6} \right)^3

\bf W(3) = 3*3*\left( \frac{5}{6} \right)^3

and the formula for the expected money won with n dice would be

\bf \boxed{W(n) = 3*n*\left( \frac{5}{6} \right)^n\;dollars}

In the picture attached there is a plot of the values of the expected money won for n=1 to 20 (See picture)

For n=5 and n=6 we get the maximum profit expected of $6.02816=$6 rounded to the nearest integer.

Hence, the smallest n that maximizes the expected payoff is n=5.

b)

The probability that face k (k=1,2,...or 6) shows up is 1/6,

as this face can be in any of the 10 positions of the arrangement, there are 10 ways that face k can show up.

The probability that face k (k=1,2,...or 6) shows up twice is \bf \left( \frac{1}{6} \right)^2

as this face can be in any of the \bf C(10;2)=\binom{10}{2} (combinations of 10 taken 2 at a time) positions of the arrangement, there are C(10;2) ways that face k can show up twice.

The probability that face k (k=1,2,...or 6) shows up three times is \bf \left( \frac{1}{6} \right)^3

as this face can be in any of the \bf C(10;3)=\binom{10}{2} (combinations of 10 taken 2 at a time) positions of the arrangement, there are C(10;3) ways that face k can show up twice.

So, we infer that the expectation is  

\bf \sum_{k=1}^{10}\binom{10}{k}(1/6)^k=3.6716\approx 4

and the expected number of distinct dice values that show up is 4.

4 0
3 years ago
(-3x-7)+(4x+7)<br><br> 2(x+14)+(2x-14)<br><br> (11x-8)+7(x-1)
inn [45]
Guessing you want to simplify each,
first one goes to just x
second goes to 4x
third goes to 18x-15

3 0
3 years ago
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