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Archy [21]
3 years ago
10

On average students attend school 170 of the 180 days required in a year. About what part of the required days do they attend sc

hool?
Mathematics
2 answers:
Llana [10]3 years ago
3 0
94% of the required days, if you wanted percentages
Delicious77 [7]3 years ago
3 0
94% of the required days, if you wanted percentagesBRAINLIEST???

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Solve for v 14v-8v=24
Vaselesa [24]
<span>14v-8v=24
Subtract 8v from 14v
6v=24
Divide 24 by 6
Final Answer: v = 4</span>
6 0
3 years ago
Read 2 more answers
What is the inverse of the given coordinates: (-2, 2) (3, 0) (7,7) *
Nana76 [90]

Answer:

2. - 2 \\  - 3.0 \\  - 7. - 7

8 0
2 years ago
Given that x/1 = 3y/8, find the ratio of : y
brilliants [131]

Answer:

3 : 8

Step-by-step explanation:

\frac{x}{1} = \frac{3y}{8}

cross multiply

8x = 3y

x = 3/8

x : y = 3 : 8

5 0
3 years ago
In a company 85% of the workers are men if 510 women work for the company how many workers are there in all
tester [92]

Answer:

3400 workers

Step-by-step explanation:

85% of men that means 15% are women

15% = 510

15/100= 510/x

100x510=51000

51000/15 =3400

x=3400

3 0
3 years ago
Components of a certain type are shipped to a supplier in batches of ten. Suppose that 50% of all such batches contain no defect
ad-work [718]

Answer:

a) P (0 defective component) = 0.5

   P( 1 defective component ) = 0.35

    P( 2 defective component ) = 0.15

b) P( 0 ) = 0

   p ( 1 ) =  1

   p ( 2 ) = 0.33

Step-by-step explanation:

p( no defective ) = 0.5

p( 1 is defective ) = 0.35

p( 2 is defective ) = 0.15

Given that 2 components are selected at random

<u>a) Given that neither component is defective </u>

Probability of 0 defective component = 0.5

P( 1 defective component ) = 0.35

P( 2 defective component ) = 0.15

<u>b) Given that one of the two tested component is defective </u>

P( 0 defective ) = 0

P( 1 defective ) = P ( \frac{x=1}{x\geq 1} ) = p( x = 1 ) / 1 - P ( x = 0 )

                                        = ( 0.5 )^1 ( 0.5 )^0 /  1 - ( 0.5)^0 (0.5)^1

                                        = 0.5 * 1 / 1 - 0.5 = 0.5 / 0.5 =  1

p ( 2 defective ) = p( x = 3 ) / 1 - P ( x = 0 )

                         = ( 0.5 )^2 ( 0.5 )^0 /  1 - ( 0.5)^0 (0.5)^2

                         = 0.25 / 0.75 = 0.33

<u />

5 0
2 years ago
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