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densk [106]
3 years ago
14

A cupboard contains 5 pairs of shoes, each of a different style. How many ways are there to select 4 of the shoes from the cupbo

ard so that the selection contains exactly one matching pair? Note that each pair of shoes consists of distinguishable left and right foot.
Mathematics
1 answer:
Andreas93 [3]3 years ago
4 0
The solution to the problem is as follows:

If there are 6 pairs of shoes:

C(6,1) = choose the matching pair = 6 choices. 
C(10,1) = 1 shoe from the other 10. 
Now you cannot choose the mate of that one, 
so you have 8 other shoes to choose from. 

6 * 10 * 8 = 480 choices of "4 shoes". 
But then we must divide by 2, because the last two shoes 
could be chosen in either order. 

That gives 240 choices - same as your teacher's answer, 
<span>but via a different route. 
</span>

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
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Identify terms, coefficients, factors, and the constant: 5(2x + 4) + 3x
KATRIN_1 [288]

Answer:

Terms:

<em>5, 2x, 4, 3x</em>

Coefficients:

<em>2 in 2x, and 3 in 3x</em>

Constants:

<em>5 and 4</em>

Factors:

<em>5, 2x, and 4</em>

Step-by-step explanation:

For terms, all of them are terms because they are all in the expression.

For coefficients, coefficients are ones that have a variable (think of it as a <em>copilot</em> kind of thing).

For factors, factors are terms that are multiplied together together to get a <em>product</em> (the answer).

For constants, these are terms that stand alone, by themselves. They are not attached to variables.

The answer is <em>13x + 20, </em>too, just in case you needed that.

Have a great day and hope this helps!

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3 years ago
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It is shown that he goes down in a slower time than he does going up.
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An airport shuttle charges 15$ plus 1.15$ per mile. An equation is written as y = mx + b, where y is the cost for x miles. (plea
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3 years ago
Can I get help solving this graph please?
Daniel [21]
see the attached figure with the letters

1) find m(x) in the interval A,B
A (0,100)  B(50,40) -------------- > p=(y2-y1(/(x2-x1)=(40-100)/(50-0)=-6/5
m=px+b---------- > 100=(-6/5)*0 +b------------- > b=100
mAB=(-6/5)x+100

2) find m(x) in the interval B,C
B(50,40)  C(100,100) -------------- > p=(y2-y1(/(x2-x1)=(100-40)/(100-50)=6/5
m=px+b---------- > 40=(6/5)*50 +b------------- > b=-20
mBC=(6/5)x-20

3) find n(x) in the interval A,B
A (0,0)  B(50,60) -------------- > p=(y2-y1(/(x2-x1)=(60)/(50)=6/5
n=px+b---------- > 0=(6/5)*0 +b------------- > b=0
nAB=(6/5)x

4) find n(x) in the interval B,C
B(50,60)  C(100,90) -------------- > p=(y2-y1(/(x2-x1)=(90-60)/(100-50)=3/5
n=px+b---------- > 60=(3/5)*50 +b------------- > b=30
nBC=(3/5)x+30

5) find h(x) = n(m(x)) in the interval A,B
mAB=(-6/5)x+100
nAB=(6/5)x
then 
n(m(x))=(6/5)*[(-6/5)x+100]=(-36/25)x+120
h(x)=(-36/25)x+120
find <span>h'(x) 
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6) find h(x) = n(m(x)) in the interval B,C
mBC=(6/5)x-20
nBC=(3/5)x+30
then 
n(m(x))=(3/5)*[(6/5)x-20]+30 =(18/25)x-12+30=(18/25)x+18
h(x)=(18/25)x+18
find h'(x) 
h'(x)=18/25=0.72 

for the interval (A,B) h'(x)=-1.44
for the interval (B,C) h'(x)= 0.72

<span> h'(x) = 1.44 ------------ > not exist</span>

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Therefore, the answer above False

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