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kodGreya [7K]
3 years ago
15

Apply the Distributive property to rewrite 2(6-x)=3. Choose the correct answer below? *

Mathematics
2 answers:
leonid [27]3 years ago
8 0

Answer:A is the answer.

Step-by-step explanation:

2(6-x)=3

12-2x=3

photoshop1234 [79]3 years ago
4 0

Answer:

12 -2x =3

Step-by-step explanation:

2(6-x)=3.

Distribute

2*6 -2*x = 3

12 -2x =3

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Find a unit vector parallel to and normal to the graph of f(x) at the indicated point. f(x) = sqrt(25-x^2) point (3,4)
cupoosta [38]
<span>y = sqrt(25-x^2) at point (3,4)

The derivative gives us the slope at 3 to be:
          -2x
 ------------ at x=3: -3/4
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</span><span>so we have a vector that is parallel to the slope of the tangent line is: <4,-3>
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<span>the mag = 5 so; unit tangent = <4/5 , -3/5>
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<span>since perpendicular lines have a -1 product between slopes we get the normal to be... <3/5,4/5>
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<span>It is <4,-3> because it is rise over run. Rise is y component of vector and run is x component of vector.</span>
4 0
3 years ago
Farrah borrowed $155 from her brother. She has already paid back $15. She plans to pay back $35 each month until the debt is pai
Sedbober [7]
B) 35x + 15 = 155 
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3 0
3 years ago
Read 2 more answers
ZAED = 137°. What is AEB?<br> A А.<br> D<br> UJ<br> с
mario62 [17]

Answer:

AEB =43

Step-by-step explanation:

The two angles form a straight line so they add to 180 degrees

AED + AEB = 180

137+AEB = 180

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4 0
3 years ago
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A statistics textbook chapter contains 60 exercises, 6 of which are essay questions. A student is assigned 10 problems. (a) What
Scilla [17]

Answer:

a) P=0.3174

b) P=0.4232

c) P=0.2594

d) The shape of the hypergeometric, in this case, is like a binomial with mean np=1.

Step-by-step explanation:

The appropiate distribution to model this is the hypergeometric distribution:

P(X=x)=\frac{\binom{s}{x}\binom{N-s}{M-x}}{\binom{N}{M}}=\frac{\binom{6}{x}\binom{54}{10-x}}{\binom{60}{10}}

a) What is the probability that none of the questions are essay?

P(X=0)=\frac{\binom{6}{0}\binom{54}{10-0}}{\binom{60}{10}}\\\\P(X=0)=\frac{1*(54!/(10!*44!)}{60!/(10!*50!)} =\frac{2.3931*10^{10}}{7.5394*10^{10}} = 0.3174

b)  What is the probability that at least one is essay?

P(X=1)=\frac{\binom{6}{1}\binom{54}{9}}{\binom{60}{10}}\\\\P(X=1)=\frac{6*(54!/(9!*43!)}{60!/(10!*50!)} =\frac{3.1908*10^{10}}{7.5394*10^{10}} =0.4232

c) What is the probability that two or more are essay?

P(X\geq2)=1-(P(0)+P(1))=1-(0.3174+0.4232)=1-0.7406=0.2594

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