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Vitek1552 [10]
3 years ago
13

What is 7 over 12 plus 5 over 6 minus 2 over 9

Mathematics
2 answers:
alexdok [17]3 years ago
4 0
1.1944 hope it helps
Dafna1 [17]3 years ago
3 0

Step-by-step explanation:

7/12 + 5/6 - 2/9

= 21/36 + 30/36 - 8/36

= (21+30-8)/36

= 43/36 or 1 7/36.

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CaHeK987 [17]

Answer:

1= 4x^2

2= 5b^4

3= 3u^2v^2w

4= 7

5= bc^3d^6

Note: ^ stands for exponents

5 0
3 years ago
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jok3333 [9.3K]

Answer:2

Step-by-step explanation:

2

8 0
4 years ago
Use a tape diagram to show m-7=17
Ostrovityanka [42]
M-17=17
+17 +17
m = 34
4 0
3 years ago
Given the rectangles perimeter, find the unknown side lengths.
Dominik [7]

The perimeter is just adding all the sides together.

a.) 180 = 40 + 40 + x + x

180 = 80 + 2x Subtract 80 on both sides

100 = 2x Take 1/2 of 100 to get x

x= 50cm


b.) 1000 = 150 + 150 + x + x

1000 = 300 + 2x Subtract 300 on both sides

700 = 2x Take 1/2 of 700 to get x

x = 350

3 0
4 years ago
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
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