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3 years ago

13

If a red suit is drawn from an ordinary deck of cards, what is the probability that the card is a diamond?

2 answers:

user100 [1]3 years ago

8 0

The probability that the card is a diamond would be 13/52 which simplifies to 1/4. There are a total of 52 cards in a deck and for each suit thete are 13.

Rzqust [24]3 years ago

3 0

1/2 if you exclude the joker(s) lol
(If it is red cards only)

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2 years ago

3^-6x(3^3 divide 3^0)^2

Greeley [361]

Answer:

${3}^{2}$

Step-by-step explanation:

$3^{ - 6} \times (3^4 \div 3^0)^2$

$\frac{1}{ {3}^{6} } \times ( {3}^{4} \div {3}^{0} {)}^{2}$

$\frac{1}{729} \times ( {3}^{4} \div {3}^{0} {)}^{2}$

$\frac{1}{729} \times ( {3}^{4} {)}^{2}$

$\frac{1}{729} \times 6561$

$\frac{1}{729} \times (729(9))$

$9$

${3}^{2}$

<h3>Hope it is helpful...</h3>

8 0

3 years ago

SOLUTION We observe that f '(x) = -1 / (1 + x2) and find the required series by integrating the power series for -1 / (1 + x2).

Ann [662]

Answer:

Required series is:

$\int{\frac{-1}{1+x^{2}} \, dx =-x+\frac{x^{3}}{3}-\frac{x^{5}}{5}+\frac{x^{7}}{7}+.....$

Step-by-step explanation:

Given that

$f'(x) = -\frac{1}{1 + x^{2}}$ ---(1)

We know that:

$\frac{d}{dx}(tan^{-1}x)=\frac{1}{1+x^{2}}$ ---(2)

Comparing (1) and (2)

$f'(x)=-(tan^{-1}x)$ ---- (3)

Using power series expansion for $tan^{-1}x$

$f'(x)=-tan^{-1}x=-\int {\frac{1}{1+x^{2}} \, dx$

$= -\int{ \sum\limits^{ \infty}_{n=0} (-1)^{n}x^{2n}} \, dx$

$= -\sum{ \int\limits^{ \infty}_{n=0} (-1)^{n}x^{2n}} \, dx$

$=-[c+\sum\limits^{ \infty}_{n=0} (-1)^{n}\frac{x^{2n+1}}{2n+1}]$

$=C+\sum\limits^{ \infty}_{n=0} (-1)^{n+1}\frac{x^{2n+1}}{2n+1}$

$=C-x+\frac{x^{3}}{3}-\frac{x^{5}}{5}+\frac{x^{7}}{7}+.....$

as

$tan^{-1}(0)=0 \implies C=0$

Hence,

$\int{\frac{-1}{1+x^{2}} \, dx =-x+\frac{x^{3}}{3}-\frac{x^{5}}{5}+\frac{x^{7}}{7}+.....$

7 0

4 years ago