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lana66690 [7]
3 years ago
15

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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PLZZ HELP URGENT
Mashcka [7]

Answer:

5x + 2x.....combine like terms..... = 7x

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....

because 5x + 2x = 7x......and the other one is 7x - 1......so the 7x - 1 values will always be 1 number less...because ur subtracting one

Step-by-step explanation:

5 0
3 years ago
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Consider the following geometric sequence.<br><br> 2, 6, 18, 54,
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Answer:

The rule or operation in this sequence is multiplication by 3.

Step-by-step explanation:

2*3=6

6*3=18

18*3=54

and so on...

7 0
2 years ago
3x-2y=10 3x-2y=14 solve using substitution or elimination. What's the answer?
CaHeK987 [17]
I hope this helps you



-3x+2y= -10


3x-2y=14



-3x+2y+3x-2y= -10+14


0+0= 6


0=6 false it's not inconsistent




8 0
3 years ago
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A company makes 120 bags.
Nana76 [90]

Answer:

43 out of 120 have zips

Step-by-step explanation:

because

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Suppose that a box contains r red balls and w white balls. Suppose also that balls are drawn from the box one at a time, at rand
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}}\\\\

7 0
3 years ago
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