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STatiana [176]
3 years ago
8

A building contractor buys 60​% of his cement from supplier A and 40​% from supplier B. A total of 95​% of the bags from A arriv

e​ undamaged, and a total of 75​% of the bags from B arrive undamaged. Find the probability that an undamaged bag is from supplier A.
Mathematics
1 answer:
docker41 [41]3 years ago
7 0

Answer: 0.665.

Step-by-step explanation:

60% of the bags are from supplier A, and 95% of them are undamaged.

40% of the bags are from supplier B, and 75% of them are undamaged.

Then, the number of undamaged bags from supplier A is:

0.6*0.95 = 0.57 or 57%

from B is:

0.4*0.75 = 0.3 or 30%

This means that the percentage of undamaged bags is:

57% + 30% = 87%.

Then, the probability that an undamaged bag is from supplier A is equal to the percentage of undamaged bags that came from supplier A (57%) and the total percentage of undamaged bags (87%) this is:

P = 57%/87% = 0.655 or 65.5%

Then the probability that an undamaged bag comes from supplier A is 0.665.

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A production manager randomly sampled production lines at a factory that produces automobiles. She wanted to find out how many p
Elena-2011 [213]

Answer:

The confidence interval for the proportion of production lines that caused defects is (0.07, 0.09).

Step-by-step explanation:

A confidence interval for a population proportion is a function of the sample proportion and the margin of error.

The interval has two bounds, a lower bound and an upper bound.

The lower bound is the sample proportion subtracted by the margin of error.

The upper bound is the margin of error added to the sample proportion.

In this problem, we have that:

Sample proportion 0.08

Margin of error 0.01

0.08 - 0.01 = 0.07

0.08 + 0.01 = 0.09

The confidence interval for the proportion of production lines that caused defects is (0.07, 0.09).

5 0
3 years ago
17 POINTS!!!! List the elements in the set described. {x|x is a natural number between 9 and 17} (Use a comma to separate answer
stepladder [879]

Answer:

10, 11, 12, 13, 14, 15, 16

Step-by-step explanation:

The natural numbers are 1, 2, 3, 4, ...

The natural numbers between 9 and 17 are 10, 11, 12, 13, 14, 15, 16.

Answer: 10, 11, 12, 13, 14, 15, 16

8 0
3 years ago
Can anyone help me. Will give you all my points and brainliest:)
PilotLPTM [1.2K]

See the attachments for the answer :)

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3 years ago
When solving problems with units we sometimes ____ to get the units the same.
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Convert is your answer .-.
4 0
3 years ago
Read 2 more answers
A large but sparsely populated county has two small hospitals, one at the south end of the county and one at the north end. The
Readme [11.4K]

Answer:

Step-by-step explanation:

Probability distribution of this type is a vicariate distribution because it specifies two random variable X and Y. We represent the probability X takes x and Y takes y by

f(x,y) = P((X=x,Y=y) ,

and given that the random variables are independent the joint pmf isbgiven by:

f(x,y) = fX(x) .fY(y) and this gives the table (see attachment)

(b) the required probability is given by considering X=0,1 and Y =0,1

f(x,y) = sum{(P(X ≤ 1, Y ≤ 1) }

= [0.1 +0.2 ] × [0.1 + 0.3]

= 0.3 × 0.4

= 0.12

Now we find

fX(x) = sum{[f(x.y)]} for y =0, 1 and similarly for fY(y)= sum{[f(x,y)]} for x=0, 1

fX(x) = sum{[f(x,y)]}= 0.1+0.3

= 0.4

fY(y) = sum{[f(x.y)]} = 0.1 + 0.2

= 0.3

Since fY(y) × fX(x) = 0.4 × 0.3

= 0.12

Hence f(x,y) = fX(x) .fY(y), the events are independent

(c) the required event is give by: [P(X<=1, PY<=1)]

= P(X<=1) . P(Y<=1)

= [sum{[f(x.y]} over Y=0,1] × [sum{[f(x.y)]}, over X= 0, 1

= 0.4 × 0.3

= 0.12

(d) the required event is given by the idea that: either The South has no bed and the North has or the the North has no bed and the South has. Let this event be A

A =sum[(P(X = 1<= x<= 4, Y =0)] + sum[P(X =0 Y= 1 <= x <= 3)]

A = [0.02 + 0.03 + 0.02 + 0.02] + [0.03 + 0.04 + 0.02]

A = [0.09] + [ 0.09]

A = 0.18

Pls note the sum of the vertical column X=2 is 0.3

3 0
3 years ago
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