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forsale [732]
3 years ago
13

For real number a, which of the following equations are true ? Select all that apply.

Mathematics
1 answer:
trapecia [35]3 years ago
3 0

ANSWER

\lim_{x \to \: a}(x)  = a

\lim_{x \to \: a}(a)  = a

\lim_{x \to \: 5}(x)  = 5

EXPLANATION

For real number 'a',

\lim_{x \to \: a}(x)  = a

is true because we have to plug in 'a' for x.

\lim_{x \to \: a}(a)  = a

This is also true because limit of a constant is the constant.

\lim_{x \to \: 5}(4)  = 5

is false. The correct value is

\lim_{x \to \: 5}(4)  =4

\lim_{x \to \: 5}(x)  = 5

is also true because we have to substitute 5 for x.

\lim_{x \to \: a}(a)  = x

is also false

The limit should be

\lim_{x \to \: a}(a)  = a

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If you have 6 cars, but there is only room in your driveway for 3 cars, in how many ways can you arrange the cars in your drivew
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If the order does not matter, cars can be arranged in 20 ways

If an order is important, cars can be arranged in 120 ways

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\left ({n} \atop {r}} \right.)=\frac{n!}{r!(n-r)!}

Here, n=6 r=3

using formula,we get

\left ( {{6} \atop {3}} \right)=\frac{6!}{3!3!} =5*4=20

2. If an order is important, Permutation will be applicable

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∴^{6} P_{3} =\frac{6!}{3!(6-3)!}=\frac{6!}{3!}  =120

3. the probability that the three newest cars end up parked in the driveway

P=\frac{No. of possible outcomes for 3 cars}{total possibilities}

=\frac{6*4*5}{6!}

=\frac{120}{720}

=\frac{1}{6} ≈ 0.167

Hence, If the order does not matter, cars can be arranged in 20 ways

If an order is important, cars can be arranged in 120 ways

The probability that the three newest cars end up parked in the driveway is 0.167

Learn more about probability here brainly.com/question/6077878

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