Help! Answer in box is wrong

g What is the probability of being dealt exactly three of a kind (like three kings or three 7’s, etc.) in a five card hand from

Elena L [17]

Answer: 0.02257

Step-by-step explanation:

Given : Total cards in a deck = 52

Number of ways to select any 5 cards : $^{52}C_5$

Since , there are total 13 kinds of card (includes Numbers from 2 to 9 and Ace , king, queen and jack).

Of each kind , there are 4 cards.

Number of ways to select three cards in a five card hand of a single kind : $^{4}C_3\times^{48}C_2$

Number of ways to select three cards in a five card hand of a exactly three of a kind : $13\times^{4}C_3\times^{48}C_2$

Now , the required probability = $\dfrac{13\times^{4}C_3\times^{48}C_2}{^{52}C_5}$

$=\dfrac{13\times4\times\dfrac{48!}{2!46!}}{\dfrac{52!}{5!47!}}\\\\\\=\dfrac{58656}{2598960}$

$=0.022569027611\approx0.02257$

∴ The probability of being dealt exactly three of a kind (like three kings or three 7’s, etc.) in a five card hand from a deck of 52 cards= 0.02257

8 0

3 years ago

5( x + 7) - 3 ( x - 4) = 7x + 2 what is x?

S_A_V [24]

Answer:

x=9

Step-by-step explanation:

5( x + 7) - 3 ( x - 4) = 7x + 2

Distribute

5x+35 - 3x +12 = 7x +2

2x +47 = 7x+2

Subtract 2x from each side

2x+47-2x =7x -2x+2

47 = 5x+2

Subtract 2 from each side

47-2= 5x+2-2

45= 5x

Divide by 5

45/5 =5x/5

9 =x

8 0

2 years ago

Read 2 more answers

The measure of an angle in standard position is given. Find two positive angles and two negative angles that are coterminal with

Pani-rosa [81]

Notice the picture below

negative angles, are just angles that go "clockwise", namely, the same direction a clock hands move hmmm so.... and one revolution is just 2π

now, you can have angles bigger than 2π of course, by simply keep going around, so, if you go around 3 times on the circle, say "counter-clockwise", or from right-to-left, counter as a clock goes, 3 times or 3 revolutions will give you an angle of 6π, because 2π+2π+2π is 6π

now... say... you have this angle here... let us find another that lands on that same spot

by simply just add 2π to it :)

$\bf \cfrac{7}{6}+2=\cfrac{19}{6}\qquad thus\qquad \cfrac{7\pi} {6}+2\pi =\cfrac{19\pi }{6}
\\\\\\
\cfrac{19\pi }{6}\impliedby co-terminal\ angle$

now, that's a positive one
and $\bf \cfrac{7}{6}-2=-\cfrac{5}{6}\qquad thus\qquad \cfrac{7\pi} {6}-2\pi =-\cfrac{5\pi }{6}
\\\\\\
-\cfrac{5\pi }{6}\impliedby \textit{is also a co-terminal angle}$

to get more, just keep on subtracting or adding 2π

8 0

3 years ago

Can someone produce SPSS results by including the the intercept (a.k.a. the constant), standard error of estimate, unstandardize

Firlakuza [10]

Answer:these are all odd

Step-by-step explanation:

the p-value for the unstandardized coefficient, R-square, Beta (i.e. standardized regression coefficient) using Excel with this given data?

Overall: 3.4, 2.9, 2.6, 3.8, 3.0, 2.5, 3.9, 4.3, 3.8, 3.4, 2.8, 2.9, 4.1, 2.7, 3.9, 4.1, 4.2, 3.1, 4.1, 3.6, 4.3, 4.0, 2.1, 3.8, 2.7, 4.4, 3.1, 3.6, 3.9, 2.9, 3.7, 2.8, 3.3, 3.7, 4.2, 2.9, 3.9, 3.5, 3.8, 4.0, 3.1, 4.2, 2.9, 3.9, 3.5, 3.8, 4.0, 3.1, 4.2, 3.0, 4.8, 3.0, 4.4, 4.4, 3.4, 4.0, 3.5

Teach: 3.8, 2.8, 2.2, 3.5, 3.2, 2.7, 4.1, 4.2, 3.7, 3.7, 3.3, 3.3, 4.1, 3.1, 2.9, 4.5, 4.3, 3.7, 4.2, 4.0, 3.7, 4.0 2.9, 4.0, 3.3, 4.4, 3.4, 3.8, 3.7, 3.1, 3.8, 3.2, 3.5, 3.8, 4.4, 3.7, 4.00, 3.4, 3.2, 3.8, 3.7, 4.3, 3.4, 4.0 3.1, 4.5, 4.8, 3.4, 4.2, 3.4

Exam: 3.8, 3.2, 1.9, 3.5, 2.8, 3.8, 3.8, 4.1, 3.6, 3.6, 3.5, 3.3, 3.6, 3.8, 3.8, 4.2, 4.1, 4.0, 4.3, 4.2, 4.0, 4.1, 2.7, 4.4, 4.4, 4.3, 3.6, 4.1, 4.2, 3.6, 4.4, 3.4, 3.2, 3.7, 4.3, 4.1, 3.7, 4.0, 3.6, 4.0, 3.7, 4.2, 4.2, 4.1, 3.2, 4.5, 4.3, 3.6, 4.00, 3.9

Knowledge: 4.5, 3.8, 3.9, 4.1, 3.5, 4.2, 4.5, 4.7, 4.1, 4.1, 3.9, 3.9, 4.0, 4.1, 4.5, 4.5, 4.5, 4.5, 4.7, 4.0, 4.5, 4.6, 3.7, 4.1, 3.6, 4.4, 3.3, 3.8, 4.2, 3.8, 4.0, 3.1, 4.4, 4.3, 5.00 4.2, 4.5, 4.5, 4.7, 4.3, 4.00, 4.2, 3.8, 4.9, 3.7, 4.6, 4.3, 3.5, 4.4, 4.4

Grade: 3.5, 3.2, 2.8, 3.3,

6 0

1 year ago

Will mark BRAINLIST for only correct answer

katovenus [111]

Answer:

Gym A

Step-by-step explanation:

A linear relationship is a relationship of the form y = mx + b, where y and x are the linear variables, m is the rate of change and b is the value of y when x = 0.

Gym A:

Let x represent the month and y represent the total cost for the gym. From the table, we can represent the values in the form (x, y) as (1,70), (2, 90) and (3, 110). We can find the relationship between x and y using the formula:

$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\\\\Let \ us\ use\ the\ points\ (1,70)\ and\ (3,110),hence:\\\\y-70=\frac{110-70}{3-1}(x-1)\\\\y-70=20(x-1)\\\\y=20x-20+70 \\\\y=20x+50\\\\For\ 10\ months(i.e.\ x=10):\\\\y=20x+50\\\\y=20(10)+50\\\\y=\$250$

Gym B:

We can represent the values from the table as (1,55), (2, 80) and (3, 105). We can find the relationship between x and y using the formula:

$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)\\\\Let \ us\ use\ the\ points\ (1,55)\ and\ (3,105),hence:\\\\y-55=\frac{105-55}{3-1}(x-1)\\\\y-55=25(x-1)\\\\y=25x-25+55 \\\\y=25x+30\\\\For\ 10\ months(i.e.\ x=10):\\\\y=25x+30\\\\y=25(10)+30\\\\y=\$28 0$

Hence gym A would cost less ($250 < $280)

4 0

2 years ago