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Paladinen [302]
2 years ago
11

tại một cửa hàng , xác suất để khách hàng mua thiết bị điện tử là 0,3, xác suất để khách hàng cài đặt phần mềm là 0,4 nhưng với

khách hàng đã mua thiết bị điện tử thì xác suất để sau đó cài đặt phần mềm là 0,7. Tính sác xuất để 1 khách hàng mua thiết bị điện tử và cài đặt phần mềm
Mathematics
1 answer:
Harrizon [31]2 years ago
8 0

Answer:

http://eldata2.neu.topica.vn/TXTOKT02/Giao%20trinh/03_NEU_TXTOKT02_Bai2_v1.0014109205.pdf

Step-by-step explanation:

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The answer is 2 and 1/8

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2 years ago
The classes at the middle school want to raise money. The sixth grade runs a bake sale for 55 hours and makes $170$170. The seve
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The sixth grade makes 3.09 per hour

The seventh graders make 2.54 per hour

the eighth graders make 2.90 per hour

So the sixth graders make the most per hour. you can check this answer by taking the amount of money made and divide it by the time that it takes to raise the total amount of money raised by each class of students.

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3 years ago
ANSWER NEEDED ASAP
eimsori [14]

Answer:

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Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
(a) How many integers from 1 through 1,000 are multiples of 2 or multiples of 9?
DENIUS [597]

Answer:

(a)There are 500 integers from 1 through 1,000 are multiples of 2.

There are 111 integers from 1 through 1,000 are multiples of 9.

(b)0.611

Step-by-step explanation:

Given the set of Integers from 1 through 1000

The least multiple of 2 is 2 and the highest multiple of 2 in the interval is 1000.

To determine the number of terms, we use the formula for the nth term of an Arithmetic Progression, since 2,4,6,... is an Arithmetic Progression,

where first term, a =2, common difference, d =2

Nth term of an A.P,

U_n= a+(n-1)d\\1000=2+2(n-1)\\1000=2+2n-2\\1000=2n\\n=500

  • There are 500 integers from 1 through 1,000 are multiples of 2.

Similarly,

Given the set of Integers from 1 through 1000

The least multiple of 9 is 9 and the highest multiple of 9 in the interval is 999.

To determine the number of terms, we use the formula for the nth term of an Arithmetic Progression, since 9,18,27,... is an Arithmetic Progression,

where first term, a =9, common difference, d =9

Nth term of an A.P,

U_n= a+(n-1)d\\999=9+9(n-1)\\999=9+9n-9\\999=9n\\n=111

  • There are 111 integers from 1 through 1,000 are multiples of 9.

(b)Probability that an integer selected at random is a multiple of 2 or a multiple of 9.

There are a total of 1000 numbers between from 1 through 1,000

n(S)=1000

n(Multiples of 2)=500

n(Multiples of 9)=111

Therefore:

P(\text{Multiples of 9}) \:OR\: P(\text{Multiples of 2})=\frac{111}{1000} + \frac{500}{1000} \\=\frac{611}{1000} \\=0.611

7 0
3 years ago
2,000 pencils are being divided into packages of 15. How many pencils will be left over?
Mashcka [7]
You have 133 pencils left.

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