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3 years ago

How many rectangles can you build with a prime number of square tiles?

1 answer:

7 0

You can only make one rectangle but that is assuming orientation does not matter and still counts as one. For example, a 3 by 1 and a 1 by 3 is the same rectangle but has different orientations so in essences there are two but really just one

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Larry gave the clerk $10 for purchase of a $3.25 note book and a $2.58 set of markers.How much q change should he receive

Xelga [282]

Larry should re receive $4.17

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3 years ago

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PLEASE HELP ASAP NO TROLLING PLEASE. HELP FIND MISSING TERM

snow_tiger [21]

Answer:

-3x is the answer

Step-by-step explanation:

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3 years ago

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Oduvanchick [21]

Answer:

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Step-by-step explanation:

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3 years ago

Three cards are drawn from a standard deck of 52 cards without replacement. Find the probability that the first card is an ace,

MrRissso [65]

Answer:

$4.82\cdot 10^{-4}$

Step-by-step explanation:

In a deck of cart, we have:

a = 4 (aces)

t = 4 (three)

j = 4 (jacks)

And the total number of cards in the deck is

n = 52

So, the probability of drawing an ace as first cart is:

$p(a)=\frac{a}{n}=\frac{4}{52}=\frac{1}{13}=0.0769$

At the second drawing, the ace is not replaced within the deck. So the number of cards left in the deck is

$n-1=51$

Therefore, the probability of drawing a three at the 2nd draw is

$p(t)=\frac{t}{n-1}=\frac{4}{51}=0.0784$

Then, at the third draw, the previous 2 cards are not replaced, so there are now

$n-2=50$

cards in the deck. So, the probability of drawing a jack is

$p(j)=\frac{j}{n-2}=\frac{4}{50}=0.08$

Therefore, the total probability of drawing an ace, a three and then a jack is:

$p(atj)=p(a)\cdot p(j) \cdot p(t)=0.0769\cdot 0.0784 \cdot 0.08 =4.82\cdot 10^{-4}$

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3 years ago

15. What is the dependent and independent variable in this scenario?

Montano1993 [528]

The dependent variable is 5 and the independent is 15

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3 years ago