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3 years ago

Is -2.9 a rational or integer or whole or natural number

Mathematics

1 answer:

Burka [1]3 years ago

7 0

Answer:

Step-by-step explanation:

It might help to get artsy/crafty with this. I made it out of construction paper, but I'm going to tell you how to draw it.

Draw a large circle, over have the page. Every number, ever, even the numbers in the smaller circles, go into this circle, fractions, decimals, negatives, zero, whole numbers, every number. Negative infinity to positive infinity and everything in between. These are rational numbers.

Now draw a smaller circle inside that one. In the smaller circle there are positive and negative numbers, but no fractions. Every number in this circle is also in the larger circle, but not every number in the bigger circle in in this smaller one. Negative infinity to positive infinity, no fractions. These are integers.

Draw a smaller circle inside the smaller one. No negative numbers in this one, only positive numbers and zero. 0 to infinity, no fractions. These are whole numbers.

One more circle inside the smallest one. Only positive whole numbers in here. No zero. 1 to infinity, no fractions. These are natural numbers.

-2.9 is negative and decimal, so not a natural, not a whole, not an integer. That leaves rational.

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2 years ago

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$\dfrac{\binom43}{\binom{52}3}=\dfrac4{22,100}=\dfrac1{5525}\approx0.00018$

In case you're asked about the probability of drawing a 3 or a 5 (and NOT three 5s), then there are 8 possible cards (four each of 3 and 5) that interest you, with a probability of $\frac8{52}=\frac2{13}\approx0.15$ of getting drawn.

b. Similar to the second case considered in part (a), there are now 12 cards of interest with a probability $\frac{12}{52}=\frac3{13}\approx0.23$ of being drawn.

c. There are four 6s in the deck, and thirteen diamonds, one of which is a 6. That makes 4 + 13 - 1 = 16 cards of interest (subtract 1 because the 6 of diamonds is being double counted by the 4 and 13), hence a probability of $\frac{16}{52}=\frac4{13}\approx0.31$ .

- - -

Note: $\binom nk$ is the binomial coefficient,

$\dbinom nk=\dfrac{n!}{k!(n-k)!}={}_nC_k=C(n,k)=n\text{ choose }k$

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