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3 years ago

X y

Mathematics

2 answers:

yKpoI14uk [10]3 years ago

8 0

Answer:

Step-by-step explanation:

koban [17]3 years ago

7 0

Answer:

ur answer

Step-by-step explanation:

The relationship between x and y is following-

x = shows the table of 2

x = shows the table of 2y = shows the table of 6

because,

x y

2*1 = 2 6*1 = 6

2*1 = 2 6*1 = 62*2 = 4. 6*2 = 12

2*1 = 2 6*1 = 62*2 = 4. 6*2 = 122*3 = 6. 6*3 = 18

2*1 = 2 6*1 = 62*2 = 4. 6*2 = 122*3 = 6. 6*3 = 182*4 = 8. 6*4 = 24

and also,

All the answer of multiplication are divisible by 2.

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The probability of a successful optical alignment in the assembly of an optical data storage product is p = 0.7. Assume the tria

zhannawk [14.2K]

Answer:

(a) P(X = 4) = 0.0189

(b) P(X $\leq$ 4) = 0.9919

(c) P(X $\geq$ 4) = 0.027

Step-by-step explanation:

We are given that the probability of a successful optical alignment in the assembly of an optical data storage product is p = 0.7 .

Since we have to find the probabilities for 1st successful alignments, so the probability distribution that we will use here is Geometric distribution.

Geometric distribution is used when we are interested in knowing the chances of our first success.

The probability distribution of geometric distribution is given by;

$P(X =x) = p(1-p)^{x-k} ; x = 1,2,3,....$

where, $x$ = number of trials

k = first success = 1

p = probability of getting success = 0.70

So, X ~ Geo(p = 0.7)

(a) Probability that the 1st successful alignment requires exactly 4 trials is given by = P(X = 4)

Here, $x$ = 4, p = 0.7 and k = 1

So, P(X = 4) = $0.7(1-0.7)^{4-1}$ = $0.7 \times 0.3^{3}$ = 0.0189

(b) Probability that the 1st successful alignment requires at most 4 trials is given = P(X $\leq$ 4)

P(X $\leq$ 4) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

= $0.7(1-0.7)^{1-1} + 0.7(1-0.7)^{2-1}+0.7(1-0.7)^{3-1}+0.7(1-0.7)^{4-1}$

= $0.7 \times 0.3^{0}+0.7 \times 0.3^{1}+0.7 \times 0.3^{2}+0.7 \times 0.3^{3}$

= 0.7 + 0.21 + 0.063 + 0.0189 = 0.9919

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

P(X $\geq$ 4) = 1 - P(X = 1) - P(X = 2) - P(X = 3)

= $1-0.7(1-0.7)^{1-1} + 0.7(1-0.7)^{2-1}+0.7(1-0.7)^{3-1}$

= $1-0.7 \times 0.3^{0}+0.7 \times 0.3^{1}+0.7 \times 0.3^{2}$

= 1 - 0.7 - 0.21 - 0.063 = 0.027

8 0

3 years ago

Ross paid $154.48 for 8 DVDs that were all the same price which is the best estimate of the cost of each DVD

elena-14-01-66 [18.8K]

Answer:

$19.36

Step-by-step explanation:

154.48/8

19.36

7 0

3 years ago

Solve the riddle.

Elena-2011 [213]

Answer:

✔ 9

Step-by-step explanation:

Assignment Egde 2021

Virtual School bouta end yall. Just gotta survive the end of the school year

Brainly?

4 0

3 years ago

Line K is parallel to line L

SpyIntel [72]

Answer:

The correct answer is A) Angle 1.

Step-by-step explanation:

We know this because they are opposite of one another. The lines don't even have to be parallel for this to be true.

4 0

3 years ago

Read 2 more answers

(a) How many integers in the range 1 through 120 are integer multiples of 2, 3, or 5? keyboard_arrow_down Solution (b) How many

Umnica [9.8K]

Answer:

(a) 88 integers

(b) 92 integers

Step-by-step explanation:

(a) integers whose last digits are divisible by 2 are multiples of two or numbers whose digits ends with zero. So for number 1-120 , all the even numbers which are sixty in number are are multiples of two.

For 3, numbers whose digits sum is divisible by three are multiples of three. 3,6,9,12,15,18,21,24,27,30 are multiples of three from numbers 1-30. we have four 30s in 120. which means numbers of integers will be 10*4 = 40integers. However out of these numbers , half are also integers of 2 which reduces the number added to 20integers.

For 5, numbers whose digits ends with 5 or 0 are multiples of 5. this gives us 24 integers for 1-120. but out of these 24integers, 16 are common integers of 2 and 3 which reduces the number added to 8integers.

Thus from 1-120 the intergers of 2,3 or 5 = 60+20+8 = 88integers.

(b) if we are considering from numbers 1-140;

for 2 we wil have 70 integers,

for 5 we will have 28 integers, but those integers that end with 0 are also integers of 2 which reduces the number added to 14.

For 7, numbers 7,14,21,28,35,42,49,56,63,70 are multiples of three from 1-70. This pattern is repeated from number 71-140. hence we have 20 integers in all. However 12 of the multiples are also multiples of either 2 or 5 which reduces the number to 8 integers.

Thus from 1-140, the integers of 2, 5, or 7 = 70+14+8 = 92integers

3 0

3 years ago