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alexandr402 [8]
3 years ago
9

4:54am to 10:08am how many hours?

Mathematics
1 answer:
NeTakaya3 years ago
7 0
5 hours and 14 minutes
You might be interested in
Three times each day, a quality engineer samples a component from a recently manufactured batch and tests it. Each part is class
Zielflug [23.3K]

Answer:

Step-by-step explanation:

Hello!

Three samples of components manufactured are taken per day. They are classified as:

D: "Conforming (suitable for its use)"

E: "Downgraded (unsuitable for the intended purpose but usable for another purpose)"

F: "Scrap (not usable)"

This classification includes the three events that may occur in your sample space S.

The experiment consists in recording the categories of the three parts tested in a day.

a. List the 27 outcomes in the sample space.

The possible outcomes in the space sample are the combinations of the three events. To avoid using the same letters as in the following questions I've named the evets as D, E, and F

S={DDD, DED, DFD, DEF, DFE, DEE, DFF, DDE, DDF , EDE, EEE, EFE, EED, EEF, EDF, EFD, EDD, EFF , FDF, FEF, FFF, FFE, FFD, FDE, FED, FDD, FEE}

b. Let A be the event that all the parts fall into the same category. List the outcomes in A.

  • A: "All the parts fall into the same category"

You have three possible outcomes for this event, that the three compounds are conforming, "DDD", that the three are unconforming, "EEE", or that the three compounds are scrap, "FFF". There are only three possible outcomes for this event.

S={DDD, EEE, FFF}

c. Let B be the event that there is one part in each category. List the outcomes in B.

  • B: "There is a part in each category"

This means, for example, The first one is conforming "D", the second one is unconforming "E" and the third one is scrap "F", then the first one may be unconforming "E", the second one is conforming "D" and the thirds one is scrap "F", and so on, you have 6 possible outcomes for this event:

S={DEF, DFE, EDF, EFD, FDE, FED}

d. Let C be the event that at least two parts are conforming. List the outcomes in C.

  • C: "At least two parts are conforming"

For this event, you can have two of the compounds to be considered conforming or the three of them.

S={DDD, DED, DFD, DDE, DDF , EDD, FDD}

A total of 7 combinations fit this event.

e. List the outcomes in A ∩ C

A ∩ C is an intersection between the event A and C, this means that there must be outcomes that are shared by both events.

Possible outcomes for A: S={DDD, EEE, FFF}

Possible outcomes for C: S={DDD, DED, DFD, DDE, DDF , EDD, FDD}

As you can see there is only one possible outcome shared by these two events. So the possible outcomes for A ∩ C are:

S= {DDD}

f. List the outcomes in A U B

A U B is the union between these two events, to see what outcomes this union has you have to add every outcome of A plus every outcome of B minus the possible outcomes that A and B share:

Possible outcomes for A: S={DDD, EEE, FFF}

Possible outcomes of B: S={DEF, DFE, EDF, EFD, FDE, FED}

The possible outcomes for A U B are:

S={DDD, EEE, FFF, DEF, DFE, EDF, EFD, FDE, FED, EEF, EDF, EFD, }

g. List the outcomes in A ∩ c

c is the complementary event of C, it could also be symbolized as C^c

If C: "At least two parts are conforming" then its complemental event will be

  • C^c: At most one part is conforming"

This means that one or none parts are conforming, and it's possible outcomes are:

S= {DEF, DFE, DEE, DFF, EDE, EEE, EFE, EED, EFF , FDF, FEF, FFF, FFE, FFD, FDE, FED, FEE}

Possible outcomes for A: S={DDD, EEE, FFF}

As you see there are two events on "A" that also appear in the definition of "C^c"

The possible outcomes for A ∩ C^c are:

S= {EEE, FFF}

h. List the outcomes in Ac ∩ C

Ac is the complementary event of A, also symbolized as A^c

If A: "All the parts fall into the same category", then its complemental event will be

  • A^c: "Not all the parts fall into the same category"

and its possible outcomes are the remaining 24 occurrences:

S={DED, DFD, DEF, DFE, DEE, DFF, DDE, DDF , EDE, EFE, EED, EEF, EDF, EFD, EDD, EFF , FDF, FEF, FFE, FFD, FDE, FED, FDD, FEE}

Possible outcomes for C: S={DDD, DED, DFD, DDE, DDF , EDD, FDD}

As you can see the events A^c and C share 6 occurrences in common, so the possible outcomes for the intersection will be:

S= {DED, DFD, DDE, DDF , EDD, FDD}

i. Are events A and C mutually exclusive? Explain.

Two events are mutually exclusive when the occurrence of one of them keeps the other from occurring, i.e. it can happen one or the other but not both.

A and C are not mutually exclusive since the three pieces may have the same category, "D: conforming" and at least two of them to be conforming "DD-" at the same time in the occurrence "DDD" and mutually exclusive events never happen at the same time.

ii. Are events B and C mutually exclusive? Explain.

B and C are mutually exclusive, you can easily see this if you compare the possible outcomes of both events:

Possible outcomes of B: S={DEF, DFE, EDF, EFD, FDE, FED}

Possible outcomes for C: S={DDD, DED, DFD, DDE, DDF , EDD, FDD}

There are no shared elements by these events. This means that if you were to take three pieces randomly sampled in one day and fit the definition of B, then they will not fir the definition of C and vice versa.

I hope it helps!

4 0
3 years ago
Help please hurry im stupid :'( will mark brainlyest
bixtya [17]
19.) Least to greatest:
Luxembourg
Singapore
Australia
Egypt
Brazil
5 0
3 years ago
A survey reports that 67% of college students prefer to drink more coffee during the exams week. If we randomly select 80 colleg
Akimi4 [234]

Answer:

The probability that at most 50 say that they drink coffee during exam week is 0.166.

Step-by-step explanation:

The random variable <em>X</em> can be defined as the number of college students who prefer to drink more coffee during the exams week.

The probability of the random variable <em>X</em> is <em>p</em> = 0.67.

A random sample of <em>n</em> = 80 college students are selected.

The response of every students is independent of the others.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 80 and <em>p</em> = 0.67.

But the sample selected is too large and the probability of success is close to 0.50.

So a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

  1. np ≥ 10
  2. n(1 - p) ≥ 10

Check the conditions as follows:

np=80\times 0.67=53.6>10\\n(1-p)=80\times (1-0.67)=26.4>10

Thus, a Normal approximation to binomial can be applied.

X\sim N(np, np(1-p))

The mean of the distribution of <em>X</em> is:

\mu=np=80\times 0.67=53.6

The standard deviation of the distribution of <em>X</em> is:

\sigma=\sqrt{np(1-p)}=\sqrt{80\times 0.67\times (1-0.67)}=4.206

A Normal distribution is a continuous distribution. So, the probability at a point cannot be computed for the Normal distribution. To compute the probability at a point we need to apply the continuity correction.

Compute the probability that at most 50 say that they drink coffee during exam week as follows:

Apply continuity correction:

P(X\leq 50)=P(X

                 =P(X

*Use a <em>z</em>-table for the probability.

Thus, the probability that at most 50 say that they drink coffee during exam week is 0.166.

5 0
3 years ago
PLEASE ANSWERRRRRRRRRR
mash [69]

Answer: OPTION B.

Step-by-step explanation:

Given the following System of equations:

\left \{ {{3x-5y=19} \atop {x+y=1}} \right.

You can use the Elimination Method to solve it. The steps are:

1. You can mutliply the second equation by -3.

2. Then you must add the equations.

3. Solve for the variable "y".

Then:

\left \{ {{3x-5y=19} \atop {-3x-3y=-3}} \right.\\.....................\\-8y=16\\\\y=\frac{16}{-8}\\\\y=-2

4. Now that you know the value of the variable "y", you must substitute it into any original equation.

5. The final step is to solve for "x" in order to find its value.

Then:

x+(-2)=1\\\\x=1+2\\\\x=3

Therefore, the solution is:

(3,-2)

8 0
4 years ago
Calculate if C=24 <br>PLEASE HELP!!! ​
neonofarm [45]

Answer:

-51

Step-by-step explanation:

24 / 8 - 6 - 48 =

3 - 6 - 48 =

-3 - 48 = -51

Hope that helps!

6 0
3 years ago
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