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2 years ago

Consider a standard deck of 52 playing cards with 4 suits.

Mathematics

2 answers:

Zanzabum2 years ago

8 0

Answer:

3/26

Step-by-step explanation:

In a deck of 52 cards, there are two red suits: Diamonds and Hearts

There are 6 red face cards:

2 red King

2 red Queen

2 red Jack

P(red face card) = 6/52 = 3/26

Lorico [155]2 years ago

6 0

Answer:

3/26

Step-by-step explanation:

The face cards are: Jack, Queen, and King. There are only three face cards for each of the 4 suits.

Among the 4 suits, two of them are red: diamonds and hearts. And, each of these two has 3 faces. That means that in a deck of 52 cards, there are 2 * 3 = 6 cards that are both red cards and face cards.

Probability is (# times specific event can occur) / (# times any general event will occur).

Here, the specific event is drawing a card that is both red and a face card (of which there are 6 ways), and the general event is drawing a card (of which there are 52 ways):

P(draw a card that is both red and a face card) = 6/52 = 3/26

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Leader is a stockbroker 14% commission each week last week she said $7200 worth of stocks how much did she make last week interm

Feliz [49]

Answer:

$1,008 weekly, $52,416 annually

Step-by-step explanation:

14% of $7200 is 7200 multiplied by .14 (% is decimal multiplied by 100)

.14 of 7200 = 1008

Since you specified the year, I'm going to assume that you need the annual salary based on the weekly salary that she gets from commissions.

$1008 per week, 52 weeks in a year

52 * 1008 = $52416, her commission in 2011

4 0

2 years ago

The Tennessee Titans scored 22 points in week 5 and 35 points in week 6. Find the percent of increase.

Shalnov [3]

Answer:

59.09%

Step-by-step explanation:

35-22=13

13/2= .59090909=59.09%

3 0

2 years ago

Solve for x 5 x − 9 = 3 x + 3

Musya8 [376]

5x-9 = 3x+3

2x-9 = 3

2x = 12

x= 6

plz mark branliest, hopoe this helps :)

7 0

3 years ago

If X and Y are independent continuous positive random

Leni [432]

a) $Z=\frac XY$ has CDF

$F_Z(z)=P(Z\le z)=P(X\le Yz)=\displaystyle\int_{\mathrm{supp}(Y)}P(X\le yz\mid Y=y)P(Y=y)\,\mathrm dy$

$F_Z(z)\displaystyle=\int_{\mathrm{supp}(Y)}P(X\le yz)P(Y=y)\,\mathrm dy$

where the last equality follows from independence of $X,Y$ . In terms of the distribution and density functions of $X,Y$ , this is

$F_Z(z)=\displaystyle\int_{\mathrm{supp}(Y)}F_X(yz)f_Y(y)\,\mathrm dy$

Then the density is obtained by differentiating with respect to $z$ ,

$f_Z(z)=\displaystyle\frac{\mathrm d}{\mathrm dz}\int_{\mathrm{supp}(Y)}F_X(yz)f_Y(y)\,\mathrm dy=\int_{\mathrm{supp}(Y)}yf_X(yz)f_Y(y)\,\mathrm dy$

b) $Z=XY$ can be computed in the same way; it has CDF

$F_Z(z)=P\left(X\le\dfrac zY\right)=\displaystyle\int_{\mathrm{supp}(Y)}P\left(X\le\frac zy\right)P(Y=y)\,\mathrm dy$

$F_Z(z)\displaystyle=\int_{\mathrm{supp}(Y)}F_X\left(\frac zy\right)f_Y(y)\,\mathrm dy$

Differentiating gives the associated PDF,

$f_Z(z)=\displaystyle\int_{\mathrm{supp}(Y)}\frac1yf_X\left(\frac zy\right)f_Y(y)\,\mathrm dy$

Assuming $X\sim\mathrm{Exp}(\lambda_x)$ and $Y\sim\mathrm{Exp}(\lambda_y)$ , we have

$f_{Z=\frac XY}(z)=\displaystyle\int_0^\infty y(\lambda_xe^{-\lambda_xyz})(\lambda_ye^{\lambda_yz})\,\mathrm dy$

$\implies f_{Z=\frac XY}(z)=\begin{cases}\frac{\lambda_x\lambda_y}{(\lambda_xz+\lambda_y)^2}&\text{for }z\ge0\\0&\text{otherwise}\end{cases}$

and

$f_{Z=XY}(z)=\displaystyle\int_0^\infty\frac1y(\lambda_xe^{-\lambda_xyz})(\lambda_ye^{\lambda_yz})\,\mathrm dy$

$\implies f_{Z=XY}(z)=\lambda_x\lambda_y\displaystyle\int_0^\infty\frac{e^{-\lambda_x\frac zy-\lambda_yy}}y\,\mathrm dy$

I wouldn't worry about evaluating this integral any further unless you know about the Bessel functions.

6 0

3 years ago

The question is shown in the picture

Masteriza [31]

20 + 25(times the number of hours) = 282.50
You add the 20 because of the one time service fee.

Now lets work this out.
20 + 25x =282.50
25x = 262.50 (subtract both sides by 2 to get the x's alone)
x = 10.5 (divide both sides by 25 to get just x)
x = 10.5 hours

So the answer is A.

3 0

3 years ago

Read 2 more answers