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2 years ago

PLLZZ HELP ANSWER THE QUESTION

Mathematics

1 answer:

Elodia [21]2 years ago

8 0

$a)\\\lim\limits_{x\to+\infty}\ln x=+\infty\\\\\lim\limits_{x\to+\infty}|x|=+\infty\\\\\lim\limits_{x\to+\infty}x=+\infty\\\\b)\\\lim\limits_{x\to+\infty}\dfrac{1}{x}=0\\\\\\c)\\\lim\limits_{x\to+\infty}\cos x=not\ exist\\\\d)\\\lim\limits_{x\to+\infty}\dfrac{1}{x}=0\\\\\text{Answer: a).}$

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3 years ago

When circuit boards used in the manufacture of compact disc players are tested, the long-run percentage of defectives is 5%. Let

sergiy2304 [10]

Answer:

(a) P(X=3) = 0.093

(b) P(X≤3) = 0.966

(c) P(X≥4) = 0.034

(d) P(1≤X≤3) = 0.688

(e) The probability that none of the 25 boards is defective is 0.277.

(f) The expected value and standard deviation of X is 1.25 and 1.089 respectively.

Step-by-step explanation:

We are given that when circuit boards used in the manufacture of compact disc players are tested, the long-run percentage of defectives is 5%.

Let X = the number of defective boards in a random sample of size, n = 25

So, X ∼ Bin(25,0.05)

The probability distribution for the binomial distribution is given by;

$P(X=r)= \binom{n}{r} \times p^{r}\times (1-p)^{n-r} ; x = 0,1,2,......$

where, n = number of trials (samples) taken = 25

r = number of success

p = probability of success which in our question is percentage

of defectivs, i.e. 5%

(a) P(X = 3) = $\binom{25}{3} \times 0.05^{3}\times (1-0.05)^{25-3}$

= $2300 \times 0.05^{3}\times 0.95^{22}$

= 0.093

(b) P(X $\leq$ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

= $\binom{25}{0} \times 0.05^{0}\times (1-0.05)^{25-0}+\binom{25}{1} \times 0.05^{1}\times (1-0.05)^{25-1}+\binom{25}{2} \times 0.05^{2}\times (1-0.05)^{25-2}+\binom{25}{3} \times 0.05^{3}\times (1-0.05)^{25-3}$

= $1 \times 1 \times 0.95^{25}+25 \times 0.05^{1}\times 0.95^{24}+300 \times 0.05^{2}\times 0.95^{23}+2300 \times 0.05^{3}\times 0.95^{22}$

= 0.966

(c) P(X $\geq$ 4) = 1 - P(X < 4) = 1 - P(X $\leq$ 3)

= 1 - 0.966

= 0.034

(d) P(1 ≤ X ≤ 3) = P(X = 1) + P(X = 2) + P(X = 3)

= $\binom{25}{1} \times 0.05^{1}\times (1-0.05)^{25-1}+\binom{25}{2} \times 0.05^{2}\times (1-0.05)^{25-2}+\binom{25}{3} \times 0.05^{3}\times (1-0.05)^{25-3}$

= $25 \times 0.05^{1}\times 0.95^{24}+300 \times 0.05^{2}\times 0.95^{23}+2300 \times 0.05^{3}\times 0.95^{22}$

= 0.688

(e) The probability that none of the 25 boards is defective is given by = P(X = 0)

P(X = 0) = $\binom{25}{0} \times 0.05^{0}\times (1-0.05)^{25-0}$

= $1 \times 1\times 0.95^{25}$

= 0.277

(f) The expected value of X is given by;

E(X) = $n \times p$

= $25 \times 0.05$ = 1.25

The standard deviation of X is given by;

S.D.(X) = $\sqrt{n \times p \times (1-p)}$

= $\sqrt{25 \times 0.05 \times (1-0.05)}$

= 1.089

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2 years ago

Solve |3k-2|=2|k+12|

Stella [2.4K]

Answer:

$k = 26$ or $k = -\dfrac{22}{5}$

Step-by-step explanation:

|3k - 2| = 2|k + 12|

$\dfrac{|3k - 2|}{|k + 12|} = 2$

$|\dfrac{3k - 2}{k + 12}| = 2$

$3k - 2 = 2(k + 12)$ or $3k - 2 = -2(k + 12)$

$3k - 2 = 2k + 24$ or $3k - 2 = -2k - 24$

$k = 26$ or $5k = -22$

$k = 26$ or $k = -\dfrac{22}{5}$

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3 years ago

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3 years ago

Read 2 more answers