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Vsevolod [243]
3 years ago
8

What conversion ratio was skipped in this multiple-step conversion? ​

Mathematics
1 answer:
dusya [7]3 years ago
3 0

Answer:

B

Step-by-step explanation:

B was missed. You have to convert this from hours into minutes before you can deal with seconds.

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What is the formula of EOQ​
vladimir1956 [14]

Answer:

Economic Order Quantity

7 0
3 years ago
Read 2 more answers
Seattle-Pipes Co. produces pipes to be supplied to a Seattle utility company. The requirement of the utility company is that the
Goshia [24]

Answer:

Step-by-step explanation:

Hello!

The study variable is

X: Pipe length.

It is known that this variable has a normal distribution and that the distribution parameter varies depending on the process used to manufacture the pipes.

Process A: μ= 200cm δ= 0.5cm

Process B: μ=201cm δ= 1cm

Process C: μ=202cm δ= 1.5cm

Pipes with a length of 200cm or more will be accepted by the utility company (X≥200), but pipes with less than 200cm length will be rejected (X<200)

a. Using Process C, you need to calculate the probability that the pipe will be rejected, symbolically:

P(X<200)

Using the distribution data of process C you have to standardize the value:

P(Z<(200-202)/1.5)= P(Z<-1.33)= 0.09176

b. The requirements change, accepting any pipe between 199 and 202, you have to calculate the probabilities of the pipes being between those lengths using the three process:

Process A:

P(199≤X≤202) = P(X≤202) - P(X≤199)

P(Z≤(202-200)/0.5)) - P(Z≤(199-200)/0.5))

P(Z≤4) - P(Z≤-2) = 1 - 0.02275 = 0.97725

The probability of the pipe being rejected is 0.02275

Process B:

P(199≤X≤202) = P(X≤202) - P(X≤199)

P(Z≤(202-201)/1)) - P(Z≤(199-201)/1))

P(Z≤1) - P(Z≤-2) = 0.84134 - 0.02275 = 0.81859

The probability of the pipe being rejected is 1-0.81859= 0.18141

Process C:

P(199≤X≤202) = P(X≤202) - P(X≤199)

P(Z≤(202-202)/1.5)) - P(Z≤(199-202)/1.5))

P(Z≤0) - P(Z≤-2) = 0.5 - 0.02275 = 0.47725

The probability of the pipe being rejected is 1-0.47725= 0.52275

The pipes manufactured using process A has fewer chances of being rejected.

c.

Process A costs $140

P(X≥200)= 1 - P(X<200)= 1 - P(Z<0)= 1 - 0.5= 0.5

Process B costs $160

P(X≥200)= 1 - P(X<200)= 1 - P(Z<-1)= 1 - 0.15866= 0.84134

Process C costs $177

P(X≥200)= 1 - P(X<200)= 1 - P(Z<-1.33)= 1 - 0.09176= 0.90824

If they where to make 100 pipes:

Using process A: 100*0.5= 50 pipes will be accepted, so they'll win 50*($200-$140)= $3000

Using process B: 100*0.84134= 84.134≅ 84 pipes will beaccepted, so they'll win 84*($200-$160)= $3360

Using the process C: 100*0.90824= 80.824≅ 90 pipes will be accepted, so they'll win 90*($200-$177)= $2070

As you can see, using process B will maximize the profits.

I hope it helps!

6 0
4 years ago
Need help ASAP! This is pretty urgent! Will mark brainliest!! please help if you can
Yuri [45]

Answer:

153.938 sq in.

Step-by-step explanation:

7 x 7 = 49

49 x pi = 153.938

pi = 3.14

7 0
3 years ago
What is the length of the third side of the right triangle
Elodia [21]
Hello there!

Answer: 20√2

Hope This Helps You!
Good Luck Studying ^-^
3 0
3 years ago
Mr .Baker has baked some muffins. If he packs them in boxes of 4, he will have 3 left over. If he packs them in boxes of 5, he w
babunello [35]

Answer:

\boxed{43}

Step-by-step explanation:

I think the easiest way to solve this problem is by brute force: trial and error.

We must find numbers that when divided by  

  • 4 leave 3 (4n + 3)
  • 5 leave 3 (5n + 3)
  • 6 leave 1  (6n + 1)

Here is a list of multiples of the integer n that satisfy the three conditions. \begin{array}{c|ccc}\mathbf{n} & \mathbf{4n+ 3} & \mathbf{5n + 3} & \mathbf{6n + 1}\\1 & 7 & 8 & 7\\2 & 11 & 13 & 13\\3 & 15 & 18 & 19\\4 & 19 & 23 & 25\\5 & 23 & 28 & 31\\6 & 27 & 33 & 37\\7 & 31 & 38 & \mathbf{43}\\8 & 35 & \mathbf{43} & 49\\9 & 39 & 48 & 55\\10 & \mathbf{43} & 53 & 61\\\end{array}\\\text{The only number that is common to all three lists is $\mathbf{43}$.}\\\text{The smallest possible number of muffins Mr. Baker could have baked is $\boxed{\mathbf{43}}$.}

Check:

43 \div 4 = 10R3\\43 \div5 = 8R3\\43 \div 6 = 7R1

OK .

6 0
3 years ago
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