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The answer is 20*sqrt(3).  I hope this helps!

we can use basic definition of probability to solve this

probability = successful outcomes / total outcomes

calculating total outcomes:

A has 3 options for group

B has 3 options for group

C has 3 options for group

so total possibilities = 3 * 3 * 3 = 27

now there are 3 total successful outcomes: One where all 3 go in first group one where all 3 go in 2nd group and one where all 3 go in 3rd group

so probability = 3/27 = \(\boxed{\frac{1}{9}} \)

ok we have the integers 0,1,2...,9

to make the sum equal to 4 we have options

4 + 0 = 4

3 + 1 = 4

2 + 2 = 4 (not possible as we cant have duplicates)

and that is all

so new we can divide this into cases.

case # 1 (4 and 0 are the largest 2 elements):

if 0 is second largest element we have no element less than it so the only subset possible is {0, 4} so 1 option

case # 2 (3 and 1 as 2 max elements):

now we obviously have 0 which we can add such that it does not mess with the max elements as any element >= 2 it would mess up the "second-largest = 1" so we have two options one with zero and one without zero {0,1,3}, {1,3}

so adding up the cases we have 3 satisfying subsets

If x and y are integers such that 12^2 \cdot 18^3 = xy, find x+y.    

12² • 18³ = xy     ——>     xy = 144 • 5832 = 839,808     

x and y could be any of the many factors of 839,808     

so let's just use the two factors that we have already    

been given ... namely, 144 and 5832     

x + y = 144 + 5832  =  5976     

This is one sum.  There could be are others possible.     

_.      

Find WX

\(U(-3,0), V(3,0), Y(-2,2), Z(2,2)\\ UY(-2.5,1)\\ m=\frac{y_Y-y_U}{x_Y-x_U}=\frac{2-0}{(-2)-(-3)}\\ m=2\)

\(f(x)=m(x-x_U)+y_U=2(x-(-3))+0\\ f(x)=2x+6\\ f(x_M)=-\frac{1}{m}(x-x_{UY})+y_{UY}=-\frac{1}{2}(x-(-2.5))+1\\ f(x_M)=-0.5x-0.25\\ x=0\\ \color{blue}M(0,-0.25)\ ,\ Center\ of\ the\ enclosing\ circle\)

\(r=\sqrt{x_Z\ ^2+(y_Z-y_M)^2}=\sqrt{2^2+(2-(-0.25))^2}\\ \color{blue}r=3.0104\ ,\ Radius\ of\ the\ enclosing\ circle\)

\(f_{circ}(x)=y_M\pm \sqrt{r^2-x^2}\\ 1=-0.25+\sqrt{3.0104^2-x^2}\\ (1+0.25)^2=3.0104^2-x^2\\ x_{WX}=\pm \sqrt{3.0104^2-1.25^2}\\ x_{WX}=\pm \sqrt{7.5}=\pm 2.7386 \\ \color{blue}\overline{WX}=5.4772 \)

 !

We will calculate each of the probabilities step by step using combinations and the structure of a standard deck of 52 playing cards.

Total possible 5-card hands:

The total number of ways to choose 5 cards out of 52 is:

(525)=2, ⁣598, ⁣960\binom{52}{5} = 2,\!598,\!960(552​)=2,598,960

(a) Probability of a Straight

A straight is 5 cards in sequence (e.g. A-2-3-4-5, 2-3-4-5-6, ..., 10-J-Q-K-A), not all the same suit (if they are, it's a straight flush). We ignore suit and order.

Step 1: Count possible ranks for a straight

There are 10 possible sequences of 5 consecutive ranks (starting with A-2-3-4-5 up to 10-J-Q-K-A).

Step 2: For each sequence, count suit combinations

Each card in the 5-card sequence can be of any of the 4 suits.

So number of ways to assign suits to the 5 cards:

45=10244^5 = 102445=1024

But we exclude straight flushes, where all 5 cards are of the same suit:
There are 4 straight flushes for each of the 10 sequences, so:

10×4=40 straight flushes10 \times 4 = 40 \text{ straight flushes}10×4=40 straight flushes

So total number of non-straight-flush straights:

10×(45−4)=10×(1024−4)=10×1020=10, ⁣20010 \times (4^5 - 4) = 10 \times (1024 - 4) = 10 \times 1020 = 10,\!20010×(45−4)=10×(1024−4)=10×1020=10,200

Final Probability:

\frac{10,\!200}{2,\!598,\!960} \approx 0.0039 \quad \text{(or about 0.39%)}

(b) Probability of a Flush

A flush is 5 cards of the same suit, not in sequence.

Step 1: Choose a suit

There are 4 suits.

Step 2: Choose any 5 cards from the 13 cards in that suit:

(135)=1287\binom{13}{5} = 1287(513​)=1287

So total flushes (including straight flushes):

4×1287=51484 \times 1287 = 51484×1287=5148

Now subtract the 40 straight flushes (from part a):

5148−40=51085148 - 40 = 51085148−40=5108

Final Probability:

\frac{5108}{2,\!598,\!960} \approx 0.001965 \quad \text{(or about 0.197%)}

(c) Probability of a Full House

A full house is 3 cards of one rank and 2 cards of another rank.

Step 1: Choose rank for the triplet

There are 13 ranks, choose 1:

(131)=13\binom{13}{1} = 13(113​)=13

Choose 3 cards of that rank from 4 suits:

(43)=4\binom{4}{3} = 4(34​)=4

Step 2: Choose rank for the pair (different from triplet)

There are 12 remaining ranks:

(121)=12\binom{12}{1} = 12(112​)=12

Choose 2 cards of that rank:

(42)=6\binom{4}{2} = 6(24​)=6

Total number of full houses:

13×4×12×6=374413 \times 4 \times 12 \times 6 = 374413×4×12×6=3744

Final Probability:

\frac{3744}{2,\!598,\!960} \approx 0.001440 \quad \text{(or about 0.144%)}

Final Answers:

(a) ~0.0039 (0.39%)
(b) ~0.001965 (0.197%)
(c) ~0.001440 (0.144%)

Let S={0,1,…,9}. We are looking for subsets of S that have at least two elements, and the sum of the two largest elements in the subset is 4.

Let the subset be A⊆S. Let the two largest elements in A be a and b, with a>b. We require a+b=4.

Let's list the possible pairs (a,b) such that a>b and a+b=4:

If a=4, then b=0. The pair is (4,0).

In this case, 4 and 0 must be in the subset.

All other elements in the subset must be less than b=0. However, there are no elements in S less than 0.

So, the only subset for this case is {4,0}. This subset has two elements, and 4+0=4. So, this is a valid subset. (1 subset)

If a=3, then b=1. The pair is (3,1).

In this case, 3 and 1 must be in the subset.

All other elements in the subset must be less than b=1. The only element in S less than 1 is 0.

So, for this pair, the subset must contain {3,1}. We can optionally include 0.

The possible subsets are:

{3,1}

{3,1,0}

Both of these subsets have at least two elements, and the two largest elements are 3 and 1, with a sum of 4. (2 subsets)

If a=2.5, not possible as elements are integers.

Notice that a must be at least 2. The largest possible value for a is 4 (since a+b=4 and b must be at least 0).

Also, b must be at least 0.

And a>b.

Let's re-examine systematically:

The two largest elements, a and b (a>b), must sum to 4.

Possible pairs (a,b) from S satisfying a+b=4 and a>b:

Case 1: a=4,b=0

The subset must contain {4,0}.

All other elements in the subset must be smaller than b=0. There are no such elements in S.

So, the only subset is {4,0}. This subset has two elements, and the sum of its two largest elements (4 and 0) is 4.

Number of subsets for this case: 1.

Case 2: a=3,b=1

The subset must contain {3,1}.

All other elements in the subset must be smaller than b=1. The only element in S smaller than 1 is 0.

So, the subset must contain {3,1}, and we can either include or exclude 0.

The possible subsets are {3,1} and {3,1,0}.

Both of these subsets have at least two elements, and the sum of their two largest elements (3 and 1) is 4.

Number of subsets for this case: 2.

We cannot have a=2 because if a=2, then b=2, but we require a>b.

Therefore, the total number of subsets satisfying the given properties is the sum of the subsets from each case:

Total subsets = (Subsets from Case 1) + (Subsets from Case 2)

Total subsets = 1 + 2 = 3.

The subsets are:

{4,0}

{3,1}

{3,1,0}

The final answer is 3​.

\(-\sqrt{3}sec(\theta)=2\\ -\sqrt{3}\cdot \dfrac{1}{cos(\theta)}=2\\ cos(\theta)=\dfrac{-\sqrt{3}}{2}\\ \theta =\dfrac{\pi}{180^{\circ}}\cdot (\arccos\dfrac{-\sqrt{3}}{2})+2\pi n\\ \theta =\dfrac{\pi}{180^{\circ}}\cdot (150^{\circ})+2\pi n\\ \color{blue}\theta=\dfrac{5\pi}{6}+2\pi n\)

Multiply the DEG-result of  \(\arccos \dfrac{-\sqrt{3}}{2} \ by\ \dfrac{\pi}{180^{\circ}}(=1)\) to express it as a fraction. To express the multiple result of \(\arccos \dfrac{-\sqrt{3}}{2}\)  add \(2\pi n.\)

 !

Firstly, let's prime factorize \({20}^{20}\):

\({20}^{20}={(4\times5)}^{20}={2}^{40}\times{5}^{20}\)

When we choose a factor of \({20}^{20}\), we are choosing a number of the form \({2}^{a}\times{5}^{b}\), where \(0 \le a \le 40\) and \(0 \le b \le 20\).

Since all divisors are equally likely to be chosen, a and b are randomized within their ranges.

If our factor is a multiple of 5, then we have \(b \ge 1\). Otherwise, \(b=0\), and our factor is not a multiple of 5.

Therefore, the probability that our factor is a multiple of 5 is the probability that \(b \ge 1\).

There are 21 numbers between 0 and 20 inclusive, and there are 20 numbers from 1 to 20 inclusive, our answer is \(\frac{20}{21}\).

What is the range of the function g(x) = (6x + 7)(2x - 3)?    

This forms a parabola that opens upward, therefore the    

range is from the vertex's y-coordinate to positive infinity.    

y  =  (6x + 7)(2x – 3)  =  12x² – 4x – 21    

y'  =  24x – 4     

set y' = 0     —>>  x = 1/6     & sub back into original    

y  =  [(12) • (1/6)²]  –  (4)(1/6)  –  21    

y  =  1/3                 – 2/3           – 21   =  – 21.3333    

The range is from – 21.3333 to positive infinity    

_.    

In triangle ABC, the measure of angle A is x degrees, the measure of angle B is 2x degrees and the measure of angle C is 6x degrees. What is the value of x? Express your answer as a decimal to the nearest tenth.    

A = x     B = 2x     C = 6x    

A + B + C  =  180^o    

x + 2x + 6x  =  180^o    

              9x  =  180^o     —>>     x = 20.0^o    

_.    

Rationalize the denominator of 5/(3 - sqrt(3)), and write your answer in the form (A + B*sqrt(3))/C where the fraction is in lowest terms. What is A + B + C?    

5 / (3 – sqrt(3))    

Multiply both top and bottom by (3 + sqrt(3))     

to get the difference of squares in the denom.    

[ 5 / (3 – sqrt(3)) ] • [ (3 + sqrt(3)) / (3 + sqrt(3)) ]    

result  5 • (3 + sqrt(3)) / (9 – 3)    

               15 + 5sqrt(3) / 6    

A = 15     B = 5     C = 6    

A+B+C = 15 + 5 + 6  =  26    

_.    

A city park commission received a donation of playground equipment from a parents' organization. The area of the playground needs to be 256 square yards for the children to use it safely. The playground will be rectangular.

In a different plan, the sides can be of any length as long as the rectangular area remains 256 square yards. What dimensions of the rectangular area provide the least perimeter of fencing?    

You get the most area for the amount of perimeter if the figure is a square.    

sqrt(256) = 16     so the dimensions would be 16 x 16    

_.    

A class of 30 students recently took a test. If 20 students scored 80, 8 students scored 90, and 2 students scored 95, then what was the class average (the mean) on this test?    

                                                       20 x 80  = 1600    

                                                         8 x 90  =   720    

                                                         2 x 95  =   190    

add them up                                                    2510    

                                                    2510 / 30  =  83.67    

_.