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LOL!!!!....not really a genius....in fact, these probability problems give many of us trouble....including me....!!!

genius.

Answered here , NSS :  http://web2.0calc.com/questions/help-please_52349#r1

A     ....since we can replace the cards....the probability that that they are both 2's is just

(4/52) * (4/52)  =  1/13 * 1/13  =   1/169

B......we are drawing without replacing....

First card a club / Second card a spade (given that the first card is a club) =

(13/52) (13/51)  =  (1/4) * (13/51)  =  13 / 204

C  The second is the dependent event .  We have the same probability for drawing a 2 both times when we replace the cards.  But without replacement, the probability on the second draw can be different.  For instance.....if we  draw a spade on the frist draw....the probability of drawing a spade on the second draw would be different  = 12/51  = 4/13. So the second draw probability depends on the first draw.

2(x + 3) > 6  or  2x + 3 ≤ −7       "or"  means  "union"....so we have

2x + 6 >  6       or      2x + 3 ≤ -7

2x > 0              or      2x  ≤ -10

x> 0                 or        x ≤ -5

This says that  

( - inf, -5 ]  or (0, inf)   ⇒   D

Let Andy's age  = x

Then Matt's age  is   twice Andy's age + 5  =  2x + 5

And we know that  

x + 2x + 5  <  35     simplify

3x + 5  < 35

3x  <  30

x < 10

So we have that       0 < x < 10    ⇒    C

Wow... thank you so much

               Truck    Car    Total

Female  5            35        40

Male      27          21        48

Total      32          56        88  

P( Buying a truck   given that he is male) 

We have 48 males......27 of them buy a truck ....so

27 / 48  = 56%

If they are independent  we must have that

P(buying a truck)  + P(being male)  =  P(buying a truck) * P(being male)

(32/ 88)   +   (48/ 88)   =    32/88 * 48/88     ????

80/88     =  24/121

90.9 %  =  19.8%   ????   not true....so these are dependent events

56%, dependent is the answer

Thanks for the answer man, really helped me out!!

-6 ≥ 10  - 4x      subtract 10 from both sides

-16 ≥ -4x         divide each side by-4.....reverse the inequality sign

4 ≤  x   ⇒   answer B

2p + 5  > 2(p -3)

2p + 5 > 2p - 6     subtract  2p from each side

5 >  - 6      this is always true.....so all real numbers are solutions    ⇒  B

thanks a bunch ツ

-3   [  1   -2    0  - 4        4 ]

              -3   15  -45   147

         ________________

         1   -5   15  -49    151

x^4  - 41x^2   = -400

x^4 - 41x^2 + 400 = 0       factor as

(x^2 - 25) (x^2 - 16)  = 0

(x + 5) (x  - 5)(x + 4) (x - 4)  = 0 

Setting each factor and solving for x produces the real roots  -5, 5, -4, 4

We have

1d^6   -  6d^5*(5y)  + 15d^4*(5y)^2  -  20d^3*(5y)^3 + 15d^2(5y)^4 - 6d*(5y)^5 + 1(5y)^6

d^6  - 30d^5y + 375d^4y^2 - 2500d^3y^3 + 9375d^2y^3 -  18750dy^5 + 15625y^6

x^4  -  6x^2  - 7x  - 6 = 0 

Using the Rational roots Theorem, 3 is a root....

Using synthetic division  to find the remaining polynomial, we have

3  [  1   0     -6      - 7     - 6  ]

            3      9        9        6

       __________________

       1   3    3         2        0

So  we have

x^3  +  3x^2  +  3x   +  2       and we can write

x^3  +  3x^2  +  2x   + x + 2     factor

x ( x^2 + 3x + 2)  +  1 (x + 2)

x (x + 2)(x + 1) +  1 ( x + 2)

(x + 2)  [ x(x + 1) + 1 ]

(x + 2) [ x^2 + x + 1]  

So -2 is the other real root

And  using the quadratic formula....the roots of the other polynomial are

[-1 ± i√3] / 2

thanks so much

10pi  ≈  31.4

So integers -31 to  -1, 0  and 1 to 31  =

31 +  1 + 31   =   

63  integers 

y = x^3

Sequentially, we have :

Vertical compression of 1/7   

y  = (1/7)x^3

Horizontal shift 8 units to the left

y  =  (1/7)(x + 8)^3

Reflection across the x axis

y  = - (1/7)(x + 8)^3

y =  ( x - 12)^3 - 7

(x - 12)^3 - 7  = 0     add 7 to both sides

(x - 12)^3  = 7        take the cube root of both sides

x  - 12  = ∛7        add 12 to both sides

x  = ∛7  + 12

Thanks!

A graph is one way to go here :

https://www.desmos.com/calculator/qweheidaya

The minimum sum  of  1 occurs when x  = 1.5

AB  =   AC   .....so  .....angles  ABD  = ACD

So....we have that

47 + 47 + 2y  =  180

94 + 2y  = 180       subtract 94 from both sides

2y  =  86      divide both sides by 2

y  = 43°

And

47 + x + y  = 180

47 + x +  43  = 180

90 + x  = 180      subtract 90 from both sides

x  = 90°

Since two sides of this triangle are equal.......the angles opposite these sides are also equal

So......we can solve this

x + x +  38  = 180

2x + 38  = 180     subtract 38 from both sides

2x  = 142      divide both sides by 2

x = 71°

8. All angles in an equilateral triangle are equal

9. The hypotenuse is always the longest side in a right triangle

10. Congruent parts of congruent triangles are themselves congruent

Example

 If ABC   congruent to  DEF   ...then......

Angle A  = angle D

Angle B  = angle E

Angle C  = angle F

AB = DE

AC = DF

BC = EF

3.  by ASA    [ Angle - Side - Angle ]

4.  CPCTC   [ Congruent Parts of Congruent Triangles are Congruent ]

Both  angles across the equal sides are themselves equal  = x

And using the exterior angle theorem,,,we have that

116  =  x  + x

116   = 2x       divide both sides by 2

58° =  x

OK..... I thought that might be the other form that you wanted....let me show you how to get this

We have two points  on the line  (0, -5)  and  ( 1,1)

The slope between these points is  [ 1 - -5 ] / [ 1 - 0 ]  =   6/1  = 6

Now ....using (0, -5)  we have that

y -  (-5)  =  6 (x - 0)

y  + 5  = 6x

y  = 6x  - 5

Thanks!