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frutty [35]
3 years ago
14

A manufacturing operations consists of 12 operations. However, five of the 12 machining operations must be completed before any

of the remaining operations can begin. Within each of these two sets, operations can be completed in any order. How many different production sequences are possible?
Mathematics
1 answer:
shutvik [7]3 years ago
4 0

Answer:

604800 possibilities

Step-by-step explanation:

There are two sets.

The first set is for the first five machining operations. This set has the following operations: A,B,C,D,E

The first operation to be made can be any of them(A,B,C,D,E). The second will be any of the four remaining ones, the third will be any of the three remaining and so on...

So for the first set there are 5*4*3*2*1 = 5! = 120 possibilities.

Now for the second set, the same logic is applied.

There will be 7*6*5*4*3*2*1 = 7! = 5040 possibilities.

Now taking into account all the 12 operations, separated by set, there will be 120*5040 = 604800 possibilities.

You can understand this calculus as that for a simple possibility of the first set, there are 5040 possibilities of the second set. The are 120 possiblities in the first set, so we have the multiplication in the paragraph above.

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Monthly sales of an SUV model are expected to increase at the rate of S′1t2 = -24t 1>3 SUVs per month, where t is time in mon
uysha [10]

The company will continue to manufacture this model for 19 months.

The question is ill-formatted. The understandable format is given below.

An SUV model's monthly sales are anticipated to rise at a rate of

S'(t)=-24^{1/3} SUVs each month, where "t" denotes the number of months and "S(t)" denotes the monthly sales of SUVs. When monthly sales of 300 SUVs are reached, the business intends to discontinue producing this model.

Find S if monthly sales of SUVs are 1,200 at time t=0 (t). How long will the business keep making this model?

Given S'(t)=-24^{1/3} ... ... (1)

Condition (1) at t=0; s(t) =1200

We find S(t)=300 then t=?

a) We find S(t)

from S'(t)=-24^{1/3}

\implies \frac{dS(t)}{dt}=-24t^{1/3} ~~[\because X'(t)=\frac{dX}{dt} \\\implies dS(t) = -24t^{1/3}dt\\

On Integrating both sides

S(t)=-18t^{4/3}+c ~...~...~(2)

Now, at t=0 then S(t)=1200

So, from (2)

1200=0+c

⇒c=1200

\thereforefrom eq (2)

S(t)=-18t^{4/3}+1200 ~...~...~(3)

b) Considering that the firm intends to cease production of this model once monthly sales exceed 300 SUVs.

So, take S(t)=300

from eq (2) 300=-18t^{4/3}+1200

t^{4/3}=\frac{1200-300}{18}\\=\frac{900}{18}=50\\\implies t=50^{3/4}\\\implies t=18.8030

Hence the company will continue 18.8020 months.

Learn more about integration here-

brainly.com/question/18125359

#SPJ10

8 0
2 years ago
Two trains leave stations 560 miles apart at the same time and travel toward each other. One train travels at 90 miles per hour
lana [24]

Answer: the two trains will meet in 2.8 hours

Step-by-step explanation:

Step 1

Speed of train A  = 90mph

Speed of train B = 110 mph

Since both trains  are travelling towards each other , their effective speed will be= 90 mph + 110mph= 200mph

Step 2

we know that Speed = distance / time

therefore Time = Distance / speed

Distance = 560 miles

Speed = 200 mph

Time = 560miles / 200mph

2.8 hours

5 0
3 years ago
Will give Branliest!!!<br> whitch inequality is true?<br><br> A) 2pi-1≤5<br> B) pi+8≤11
Tasya [4]
Neither, the first one is greater than 5 and so is 11
8 0
3 years ago
“The 7th grade class is planning a pancake breakfast. All materials are donated except the pancake mix. Based on the given infor
ArbitrLikvidat [17]

Answer:

a) 720

b) 60$

Step-by-step explanation:

a) each person is served 3 pancakes and there are 240 people, 240 x 3 = 720.

b) Each bag of mix costs 10$ and can make 120 pancakes, but we need 720 pancakes, which means six bags of mix (since 120x6 = 720). Since each cost 10$, 10 x 6 = 60$

6 0
3 years ago
According to an NRF survey conducted by BIGresearch, the average family spends about $237 on electronics (computers, cell phones
Usimov [2.4K]

Answer:

(a) Probability that a family of a returning college student spend less than $150 on back-to-college electronics is 0.0537.

(b) Probability that a family of a returning college student spend more than $390 on back-to-college electronics is 0.0023.

(c) Probability that a family of a returning college student spend between $120 and $175 on back-to-college electronics is 0.1101.

Step-by-step explanation:

We are given that according to an NRF survey conducted by BIG research, the average family spends about $237 on electronics in back-to-college spending per student.

Suppose back-to-college family spending on electronics is normally distributed with a standard deviation of $54.

Let X = <u><em>back-to-college family spending on electronics</em></u>

SO, X ~ Normal(\mu=237,\sigma^{2} =54^{2})

The z score probability distribution for normal distribution is given by;

                                 Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean family spending = $237

           \sigma = standard deviation = $54

(a) Probability that a family of a returning college student spend less than $150 on back-to-college electronics is = P(X < $150)

        P(X < $150) = P( \frac{X-\mu}{\sigma} < \frac{150-237}{54} ) = P(Z < -1.61) = 1 - P(Z \leq 1.61)

                                                             = 1 - 0.9463 = <u>0.0537</u>

The above probability is calculated by looking at the value of x = 1.61 in the z table which has an area of 0.9463.

(b) Probability that a family of a returning college student spend more than $390 on back-to-college electronics is = P(X > $390)

        P(X > $390) = P( \frac{X-\mu}{\sigma} > \frac{390-237}{54} ) = P(Z > 2.83) = 1 - P(Z \leq 2.83)

                                                             = 1 - 0.9977 = <u>0.0023</u>

The above probability is calculated by looking at the value of x = 2.83 in the z table which has an area of 0.9977.

(c) Probability that a family of a returning college student spend between $120 and $175 on back-to-college electronics is given by = P($120 < X < $175)

     P($120 < X < $175) = P(X < $175) - P(X \leq $120)

     P(X < $175) = P( \frac{X-\mu}{\sigma} < \frac{175-237}{54} ) = P(Z < -1.15) = 1 - P(Z \leq 1.15)

                                                         = 1 - 0.8749 = 0.1251

     P(X < $120) = P( \frac{X-\mu}{\sigma} < \frac{120-237}{54} ) = P(Z < -2.17) = 1 - P(Z \leq 2.17)

                                                         = 1 - 0.9850 = 0.015

The above probability is calculated by looking at the value of x = 1.15 and x = 2.17 in the z table which has an area of 0.8749 and 0.9850 respectively.

Therefore, P($120 < X < $175) = 0.1251 - 0.015 = <u>0.1101</u>

5 0
3 years ago
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