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Sunny_sXe [5.5K]
2 years ago
5

If 6j-5k =11 and 5j-6k =22, then what is the value of 2j+2k??

Mathematics
1 answer:
dsp732 years ago
8 0

Answer:

-22

Step-by-step explanation:

6j-5k=11

5j-6k=22

-----------------

5(6j-5k)=5(11)

-6(5j-6k)=-6(22)

-----------------------

30j-25k=55

-30j+36k=-132

-----------------------

11k=-77

k=-77/11

k=-7

6j-5(-7)=11

6j+35=11

6j=11-35

6j=-24

j=-24/6

j=-4

------------------

2j+2k=2(-4)+2(-7)=-8-14=-22

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Josh travels frequently. For a particular airline, it takes 20 minutes for the first bag to arrive in baggage claim after a flig
Orlov [11]

Using the uniform distribution, it is found that:

A,B) 0.3 of the time does it take longer than 27 minutes for Josh’s bag to arrive in baggage claim.

C) 20% of the time does Josh’s bag arrive in less than 22 minutes.

--------------------------

An uniform distribution has two bounds, a and b.  

The probability of finding a value of at lower than x is:

P(X < x) = \frac{x - a}{b - a}

The probability of finding a value between c and d is:

P(c \leq X \leq d) = \frac{d - c}{b - a}

The probability of finding a value above x is:

P(X > x) = \frac{b - x}{b - a}

--------------------------

  • In the graph, we have that the distribution is uniform between 20 and 30 minutes, thus a = 20, b = 30

--------------------------

Itens a and b:

  • Above 27 minutes, thus:

P(X > 27) = \frac{30 - 27}{30 - 20} = 0.3

0.3 of the time does it take longer than 27 minutes for Josh’s bag to arrive in baggage claim.

--------------------------

Item c:

  • Less than 22 minutes, thus:

P(X < 2) = \frac{22 - 20}{30 - 20} = 0.2

0.2*100 = 20%

20% of the time does Josh’s bag arrive in less than 22 minutes.

A similar problem is given at brainly.com/question/15855314

6 0
2 years ago
SAT Writing scores are normally distributed with a mean of 491 and a standard deviation of 113.A university plans to send letter
Sholpan [36]

Answer:

z=1.405

And if we solve for a we got

a=491 +1.405*113=649.765

So the value of height that separates the bottom 92% of data from the top 8% is 649.765.  

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the scores of a population, and for this case we know the distribution for X is given by:

X \sim N(491,113)  

Where \mu=491 and \sigma=113

For this part we want to find a value a, such that we satisfy this condition:

P(X>a)=0.08   (a)

P(X   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.92 of the area on the left and 0.08 of the area on the right it's z=1.405. On this case P(Z<1.405)=0.92 and P(z>0.92)=0.08

If we use condition (b) from previous we have this:

P(X  

P(z

But we know which value of z satisfy the previous equation so then we can do this:

z=1.405

And if we solve for a we got

a=491 +1.405*113=649.765

So the value of height that separates the bottom 92% of data from the top 8% is 649.765.  

6 0
2 years ago
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