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3 years ago

Help please!!! What is P(0.6 ≤ z ≤ 2.12)? 16% 26% 73% 98%

Mathematics

2 answers:

Stels [109]3 years ago

8 0

Answer:

The correct option is 2.

Step-by-step explanation:

We have to find the value of P(0.6 ≤ z ≤ 2.12).

$P(0.6\leq z\leq 2.12)=P(z\leq 2.12)-P(z< 0.6)$

From the Cumulative Standardized Normal Distribution table,

$P(z\leq 2.12)=0.9830$

$P(z$

$P(0.6\leq z\leq 2.12)=0.9830-0.7257$

$P(0.6\leq z\leq 2.12)=0.2573$

$P(0.6\leq z\leq 2.12)\approx 0.26=26\%$

Therefore the correct option is 2.

Lelu [443]3 years ago

5 0

26% is the answer to your question

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Answer:

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5x + 2x....subbing in 1 7x - 1....subbing in 1

5(1) + 2(1) = 5 + 2 = 7 7(1) - 1 = 7 - 1 = 6

5x + 2x...subbing in 2 7x - 1...subbing in 2

5(2) + 2(2) = 10 + 4 = 14 7(2) - 1 = 14 - 1 = 13

5x + 2x...subbing in 3 7x - 1...subbing in 3

5(3) + 2(3) = 15 + 6 = 21 7(3) - 1 = 21 - 1 = 20

5x + 2x...subbing in 4 7x - 1....subbing in 4

5(4) + 2(4) = 20 + 8 = 28 7(4) - 1 = 28 - 1 = 27

5x + 2x...subbing in 5 7x - 1...subbing in 5

5(5) + 2(5) = 25 + 10 = 35 7(5) - 1 = 35 - 1 = 34

5x + 2x result values are 1 more then 7x - 1 result values

there are no values that will make the 2 expressions equal....

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Step-by-step explanation:

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3x-2y=10 3x-2y=14 solve using substitution or elimination. What's the answer?

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3x-2y=14

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dybincka [34]

Answer: Part a) $P(a)=\frac{1}{\binom{r+w}{r}}$

part b) $P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}$

Step-by-step explanation:

The probability is calculated as follows:

We have proability of any event E = $P(E)=\frac{Favourablecases}{TotalCases}$

For part a)

Probability that a red ball is drawn in first attempt = $P(E_{1})=\frac{r}{r+w}$

Probability that a red ball is drawn in second attempt= $P(E_{2})=\frac{r-1}{r+w-1}$

Probability that a red ball is drawn in third attempt = $P(E_{3})=\frac{r-2}{r+w-1}$

Generalising this result

Probability that a red ball is drawn in [tex}i^{th}[/tex] attempt = $P(E_{i})=\frac{r-i}{r+w-i}$

Thus the probability that events $E_{1},E_{2}....E_{i}$ occur in succession is

$P(E)=P(E_{1})\times P(E_{2})\times P(E_{3})\times ...$

Thus $P(E)$ = $\frac{r}{r+w}\times \frac{r-1}{r+w-1}\times \frac{r-2}{r+w-2}\times ...\times \frac{1}{w}\\\\P(E)=\frac{r!}{(r+w)!}\times (w-1)!$

Thus our probability becomes

$P(E)=\frac{1}{\binom{r+w}{r}}$

Part b)

The event " r red balls are drawn before 2 whites are drawn" can happen in 2 ways

1) 'r' red balls are drawn before 2 white balls are drawn with probability same as calculated for part a.

2) exactly 1 white ball is drawn in between 'r' draws then a red ball again at $(r+1)^{th}$ draw

We have to calculate probability of part 2 as we have already calculated probability of part 1.

For part 2 we have to figure out how many ways are there to draw a white ball among (r) red balls which is obtained by permutations of 1 white ball among (r) red balls which equals $\binom{r}{r-1}$

Thus the probability becomes $P(E_i)=\frac{\binom{r}{r-1}}{\binom{r+w}{r}}=\frac{r}{\binom{r+w}{r}}$

Thus required probability of case b becomes $P(E)+ P(E_{i})$

= $P(b)=\frac{1}{\binom{r+w}{r}}+\frac{r}{\binom{r+w}{r}}\\\\$

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