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denis-greek [22]
3 years ago
11

Visually impaired students. The Journal of Visual Impairment & Blindness (May-June 1997) published a study of the lifestyles

of visually impaired students. Using diaries, the students kept track of several variables, including number of hours of sleep obtained in a typical day. These visually impaired students had a mean of 9.06 hours and a standard deviation of 2.11 hours. Assume that the distribution of the number of hours of sleep for this group of students is approximately normal. Find the probability that a visually impaired student b. obtains less than 6 hours of sleep on a typical day. Find the probability that a visually impaired student gets between 8 and 10 hours of sleep on a typical day. Twenty percent of all visually impaired students obtain less than how many hours of sleep on a typical is day?
Mathematics
1 answer:
alukav5142 [94]3 years ago
8 0

Answer:

wdwqQ

Step-by-step explanation:

WDWDQ

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Write a series of rigid motions that transform pentagon ABCDE to pentagon A′B′C′D′E′
worty [1.4K]

Answer:

We need the following three rigid motions:

i) Reflection around y-axis, ii) Translation three units in the -y direction, iii) Translation four units in the -x direction.

Step-by-step explanation:

We need to perform three operations on pentagon ABCDE to create pentagon A'B'C'D'E':

i) Reflection around y-axis:

(x',y') = (-x,y) (Eq. 1)

ii) Translation three units in the -y direction:

(x'',y'') = (x', y'-3) (Eq. 2)

iii) Translation four units in the -x direction:

(x''',y''') = (x''-4, y'') (Eq. 3)

We proceed to proof the effectiveness of operations defined above by testing point D:

1) D(x,y) = (-1, 4) Given.

2) (x',y') = (1,4) By (Eq. 1)

3) (x'',y'') = (1, 1) By (Eq. 2)

4) D'(x,y) = (-3,1) By (Eq. 3)/Result

7 0
2 years ago
What is the probability of drawing two blue marbles if the first one is placed back in the bag before the second draw
Ilia_Sergeevich [38]
Break it down into two parts. First, what is the probability of drawing a blue marble on the first draw?Since there are 5 blue marbles and 10 total, the probability is 5⁄10, or 1/2. Now since we no longer have that blue marble, there are 4 blue marbles and 9 total. The chances of drawing a blue marble are 4/9. Therefore, the chance that both marbles drawn are blue is the chance that the first one is blue times the chance that the second one is blue. 1/2 * 4/9 = 4/18 = 2/9 Remember, math is always trying to trick you. It wants you to try and do the whole big problem at once, which can be difficult. Break it down into smaller problems, then use your answers to small parts to find the answer to the big question. Hope that helps,
7 0
3 years ago
Solve -1 3/8 K equals 2/3
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Answer:

  K = -16/33

Step-by-step explanation:

For a problem like this, it is convenient to write the mixed number as an improper fraction:

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Now, multiply the equation by the reciprocal of the coefficient of K.

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2 years ago
A)Loss is denoted by a negative integer and profit is denoted by a positive integer. A
Blizzard [7]
Hi I am here to help the answer is 46-90
6 0
2 years ago
Our faucet is broken, and a plumber has been called. The arrival time of the plumber is uniformly distributed between 1pm and 7p
Ymorist [56]

Answer:

E(A+B) = E(A)+E(B)=4+0.5 =4.5 hours

Var(A+B)= Var(A)+Var(B)=3+0.25 hours^2=3.25 hours^2

Step-by-step explanation:

Let A the random variable that represent "The arrival time of the plumber ". And we know that the distribution of A is given by:

A\sim Uniform(1 ,7)

And let B the random variable that represent "The time required to fix the broken faucet". And we know the distribution of B, given by:

B\sim Exp(\lambda=\frac{1}{30 min})

Supposing that the two times are independent, find the expected value and the variance of the time at which the plumber completes the project.

So we are interested on the expected value of A+B, like this

E(A +B)

Since the two random variables are assumed independent, then we have this

E(A+B) = E(A)+E(B)

So we can find the individual expected values for each distribution and then we can add it.

For ths uniform distribution the expected value is given by E(X) =\frac{a+b}{2} where X is the random variable, and a,b represent the limits for the distribution. If we apply this for our case we got:

E(A)=\frac{1+7}{2}=4 hours

The expected value for the exponential distirbution is given by :

E(X)= \int_{0}^\infty x \lambda e^{-\lambda x} dx

If we use the substitution y=\lambda x we have this:

E(X)=\frac{1}{\lambda} \int_{0}^\infty y e^{-\lambda y} dy =\frac{1}{\lambda}

Where X represent the random variable and \lambda the parameter. If we apply this formula to our case we got:

E(B) =\frac{1}{\lambda}=\frac{1}{\frac{1}{30}}=30min

We can convert this into hours and we got E(B) =0.5 hours, and then we can find:

E(A+B) = E(A)+E(B)=4+0.5 =4.5 hours

And in order to find the variance for the random variable A+B we can find the individual variances:

Var(A)= \frac{(b-a)^2}{12}=\frac{(7-1)^2}{12}=3 hours^2

Var(B) =\frac{1}{\lambda^2}=\frac{1}{(\frac{1}{30})^2}=900 min^2 x\frac{1hr^2}{3600 min^2}=0.25 hours^2

We have the following property:

Var(X+Y)= Var(X)+Var(Y) +2 Cov(X,Y)

Since we have independnet variable the Cov(A,B)=0, so then:

Var(A+B)= Var(A)+Var(B)=3+0.25 hours^2=3.25 hours^2

3 0
3 years ago
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