4 years ago

Nancy decides to roll a die three times. What's the probability that she will roll all even numbers?

Mathematics

1 answer:

nignag [31]4 years ago

7 0

Answer:

1/8

Step-by-step explanation:

The probability of rolling an even number on one roll is 3/6 = 1/2. Assuming the rolls are independent, the probability of rolling an even number 3 times is ...

(1/2)(1/2)(1/2) = 1/8

The probability that Nancy rolls all even numbers in 3 rolls is 1/8.

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Yuki888 [10]

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3 years ago

For weeding the garden Arjun was given $18.29 . Now he has $42.38 . How much money did he have before ?

Kipish [7]

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Step-by-step explanation:

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4 years ago

Determine all real numbers a$ such that the inequality $ |x^2 2ax 3a|\le2$ has exactly one solution in x$.

vovikov84 [41]

Answer:

$x=-1, S=\{x\:\in\mathbb{R}:x=-1\}$

Step-by-step explanation:

For the sake of clarity, assuming you meant:

$|x^{2}+2x+3|\le2$

1) Absolute Value or Modulus functions has one property that assures us that:

$If |a|$

2) Solving for x

$I) x^{2}+2x+3\geqslant -2\Rightarrow x^{2}+2x+5\geqslant 0 \Rightarrow x\geqslant \frac{-2\pm \sqrt{2^{2}-4*1*5}}{2} x'\geqslant \frac{-2+4i }{2} \:and\:x''\geqslant \frac{-2-4i }{2}$

Not defined in Real Set of Numbers

Evaluating $x^{2}+2x+3\leqslant 2$ :

$x^{2}+2x+3\leqslant 2\\x^{2}+2x+1\leqslant 0\\(x+1)(x+1)\leqslant 0 \Rightarrow S=\{-1\}$

3) So, since for the first case the Discriminant Δ <0, then the solutions presented for $x^{2}+2x+3\geqslant -2\: \in \mathbb{C}$ . The only solution in the Real Set for the inequality $|x^{2}+2x+3|\le2$ is $S=\{-1\}$ , i.e. x=-1.