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2 years ago

Help! Everyone keeps trolling me!

Mathematics

1 answer:

zhenek [66]2 years ago

5 0

9514 1404 393

Answer:

0 < x < 4
(- ∞ < x < 0) ∪ (4 < x < ∞)
x ∈ {0, 4}

Step-by-step explanation:

1. The solution is the set of x-values for which the graph is above the x-axis, where y = 0. Those x-values are in the interval (0, 4).

2. The solution is the set of x-values for which the graph is below the x-axis. Those x-values are in either of the two intervals (-∞, 0) or (4, ∞).

3. The x-intercepts of the graph are x=0 or x=4.

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(a) Starting with the geometric series [infinity] xn n = 0 , find the sum of the series [infinity] nxn − 1 n = 1 , |x| < 1.

Mila [183]

Let f(x) be the sum of the geometric series,

$f(x)=\displaystyle\frac1{1-x} = \sum_{n=0}^\infty x^n$

for |x| < 1. Then taking the derivative gives the desired sum,

$f'(x)=\displaystyle\boxed{\dfrac1{(1-x)^2}} = \sum_{n=0}^\infty nx^{n-1} = \sum_{n=1}^\infty nx^{n-1}$

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2 years ago

Suppose that two openings on an appellate court bench are to be filled from current municipal court judges. The municipal court

Ksju [112]

Answer:

$(a)\dfrac{92}{117}$

$(b)\dfrac{8}{39}$

$(c)\dfrac{25}{117}$

Step-by-step explanation:

Number of Men, n(M)=24

Number of Women, n(W)=3

Total Sample, n(S)=24+3=27

Since you cannot appoint the same person twice, the probabilities are without replacement.

(a)Probability that both appointees are men.

$P(MM)=\dfrac{24}{27}X \dfrac{23}{26}=\dfrac{552}{702}\\=\dfrac{92}{117}$

(b)Probability that one man and one woman are appointed.

To find the probability that one man and one woman are appointed, this could happen in two ways.

A man is appointed first and a woman is appointed next.
A woman is appointed first and a man is appointed next.

P(One man and one woman are appointed) $=P(MW)+P(WM)$

$=(\dfrac{24}{27}X \dfrac{3}{26})+(\dfrac{3}{27}X \dfrac{24}{26})\\=\dfrac{72}{702}+\dfrac{72}{702}\\=\dfrac{144}{702}\\=\dfrac{8}{39}$

(c)Probability that at least one woman is appointed.

The probability that at least one woman is appointed can occur in three ways.

A man is appointed first and a woman is appointed next.
A woman is appointed first and a man is appointed next.
Two women are appointed

P(at least one woman is appointed) $=P(MW)+P(WM)+P(WW)$

$P(WW)=\dfrac{3}{27}X \dfrac{2}{26}=\dfrac{6}{702}$

In Part B, $P(MW)+P(WM)=\frac{8}{39}$

Therefore:

$P(MW)+P(WM)+P(WW)=\dfrac{8}{39}+\dfrac{6}{702}\\$P(at least one woman is appointed)=\dfrac{25}{117}$

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Step-by-step explanation:

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