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Mathematics

1 answer:

erica [24]3 years ago

5 0

Find the volume of the first box which is l*w*h and second box which is also l*w*h and add them. (6*4*5) + (10*10*2)

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A root of x2<br> – 5x – 1 = 0 is

My name is Ann [436]

x =(-5+√29)/2= 0.193

5 0

3 years ago

A bridge hand is made up of 13 cards from a deck of 52. find the probability that a hand chosen at random contains at least 3 ki

Vlad [161]

1.
In total there are C(52, 13) ways that we can pick a hand, that is $\frac{52!}{13!39!}$

2.
P(a hand contains at least 3 kings)
=P(a hand contains exactly 3 kings)+P(a hand contains 4 kings)

3.
first let's find P(a hand contains exactly 3 kings):

P(a hand contains exactly 3 kings)
=n(a hand contains exactly 3 kings)/C(52, 13)

n(a hand contains exactly 3 kings)=C(4, 3)*C(48, 10)

where C(4,3) is the total number of ways we can pick 3 out of 4 kings,

C(48, 10) is the number of picking 10 letters to complete a hand, out of the 52-4=48 non-king cards.

so P(a hand contains exactly 3 kings)=[C(4, 3)*C(48, 10)]/C(52, 13)

4. with the same reasoning as in step 3:

P(a hand contains 4 kings)=n(a hand contains 4 kings)/C(52, 13)

= [C(4, 4)*C(48, 9]/C(52, 13)

5.

P(a hand contains at least 3 kings)
=P(a hand contains exactly 3 kings)+P(a hand contains 4 kings)

=[C(4, 3)*C(48, 10)]/C(52, 13)+ [C(4, 4)*C(48, 9)]/C(52, 13)

= $\frac{C(4, 3)*C(48, 10)+C(4, 4)*C(48, 9)}{C(52, 13)}$

= $\frac{4* \frac{48!}{10!38!} + \frac{48!}{9!39!}}{ \frac{52!}{13!39!} }$

simplify by 38! in the denominators and 48! in the numerators :

$\frac{4* \frac{1}{10!} + \frac{1}{9!39}}{ \frac{52*51*50*49}{13!39} } 

$

now simplify by 9! in all denominators:

$\frac{ \frac{4}{10}+ \frac{1}{39} }{ \frac{52*51*50*49}{13*12*11*10*39}}$

$\frac{ 0.426 }{ 9.7} =0.044$

4 0

4 years ago

Simplify by factoring. Assume that all expressions under radicals represent nonnegative numbers.

Nimfa-mama [501]

<h2>Answer:</h2>

<u>x>0 and working with real numbers</u>

$10x^2\sqrt{2}$