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Vilka [71]
3 years ago
14

Mark tossed six balls while playing a number game. Three ball landed in one section, and three balls landed in another section.

His score is greater than one hundred thousand. What could his score be?
Mathematics
1 answer:
luda_lava [24]3 years ago
4 0
Mark's score could be 120,000. Each ball would be worth 20,000 points and since 2,000 times 6 is 120,000, that's how many points he could have. 
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Travka [436]

Answer:

(4,0)

Step-by-step explanation:

When line intercepts x axis, y must be 0

when y =0

3x-2(0)=12

x=4

hence answer is (4,0)

8 0
3 years ago
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Suppose each laptop of a certain type is assigned a series number, which consists of a sequenceof eight symbols: number, letter,
Varvara68 [4.7K]

Answer:

Step-by-step explanation:

There are 5numbers and 3 letters. Each number can be one among the 10 possible.

Each letter is 1 among the 26 available.

a)The number of possible unique serial numbers are

10^5x 26³

The number of 5 number combinations each different (order important) is 5!(10/5)= 30240

The number of 3 letter combinations each different (order important) is 3!(26/3)=15600

The required probability (that all symbols are different if one laptop is picked at random with equal probability) is

15600x3240/ 10^5x 26³=0.2684

b) The first letter must be at least 4 to be largest among the 5 numbers. So the first letter can be one of . 4,5,6,7,8,9..

The number of 5 number combinations each different and starting with 4 (order important) is .

4!=24

The number of 5 number combinations each different and starting with 5 (order important) is .

4!(5/4)= 120

The number of 5 number combinations each different and starting with 6 (order important) is .

4!(6/4)= 360

The number of 5 number combinations each different and starting with 7 (order important) is .

4!(7/4)= 840

The number of 5 number combinations each different and starting with 8 (order important) is

4!(8/4)= 1680

The number of 5 number combinations each different and starting with 9 (order important) is .

4!(9/4)= 3024

Thus, there are

The required probability ( that all symbols are different and the first number is the largest among the numbers) is

15600* 6040/10^5x 26³= 0.0537

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3 years ago
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3 years ago
Consider 8 blood donors chosen randomly from a population. The probability that the donor has type A blood is .40. Which of the
belka [17]

Answer: Option B is the only correct option.

Step-by-step explanation:

Number of samples = n = 8.

Probability of success = p = 0.4

Probability of failure = q = 0.6

r = chosen number of donors among the 8

To solve this question, we use the distribution formula

P(x=r) = nCr * p^r * q^n-r

For option A, to check if P(3<x<5) = 0.37. [3 and 5 inclusive]

When x = 3

P(x=3) = 8C3 * 0.4^3 * 0.6^5

P(x=3) = 56 * 0.064 * 0.07776

P(x=3) = 0.2787

When x= 4

P(x=4) = 8C4 * 0.4^4 * 0.6^4

P(x=4) = 70 * 0.0256 * 0.1296

P(x=4) = 0.2322

Since p(x=3) + p(x=4) is already greater than 0.37, then we know option A is NOT correct.

For option B, To check if the probability of 1 or fewer donor is about 0.11. i.e if P(x</=1) = 0.11

When x=o

P(x=0) = 8C0 * 0.4^0 * 0.6^8

P(x=0) = 1* 1 * 0.016796

P(x=0) = 0.016796.

When x = 1

P(x=1) = 8C1 * 0.4^1 * 0.6^7

P(x=1) = 8 * 0.4 * 0.02799

P(x=1) = 0.08958

P(x=0) + P(x=1) = 0.016796 + 0.08958

P(x=0) + P(x=1) = 0.10635.

Since this is approximately 0.11, then option B is a correct option.

For option C to check if the probability 7 or more donors not having type A = 0.0087

To do this,we determine thw probability of 7 or more donors having type A and we subtract our answer from 1.

First, we determine P(x>/=7)

When x= 7

P(x=7) = 8C7 * 0.4^7 * 0.6^1

P(x=7) = 8 * 0.001638 * 0.6

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P(x=8) = 8C8 * 0.4^8 * 0.6^0

P(x=8) = 1 * 0.0006554 * 1

P(x=8) = 0.0006554

P(x=7) + P(x=8) = 0.007864 + 0.0006554 = 0.00852.

Since probability of 7 or more donors having type A is 0.00852 as against what was stated in the option C, then option C is NOT a correct option.

For option D, to check if the probability of exactly 5donors having type A blood = 0.28

When x=5

P(x=5) = 8C5 * 0.4^5 * 0.6^3

P(x=5) = 56 * 0.01024 * 0.216

P(x=5) = 0.1239.

Since probability of what was derived for having exactly 5 donors having sample A is different from what wqs given in the option, then option D is NOT correct.

For option E, since what was stated in the option negates what was derived for exactly 5 donors, then option E is NOT correct

6 0
3 years ago
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disa [49]

Answer:

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