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adelina 88 [10]
3 years ago
9

You have a normal distribution of hours per week that music students practice. The mean of the values is 8 and the standard devi

ation of the values is 4. According to the normal distribution model that corresponds to this population, what percentage of the students practice between 6 to 10 hours per week? Use the graph of the standard normal distribution given below to find the percentage. 19% 30% 34% 38%
Mathematics
1 answer:
Alona [7]3 years ago
5 0

Answer:

38%

Step-by-step explanation:

Given that X,the hours per week the music student practice follow a normal distribution.

X:N(8,4)

We have to find the percentage of students who practice between 6 and 10 hours.

6<x<10 implies converting to z

We know z = (x-mean)/sigma = (x-8)/4

Hence 6<x<10 is equivalent to (6-8/4)<Z<(10-8/4)

= |z|<0.5

From std normal table we find this area equals = 0.1915+0.1915

=0.3830 = 0.38 (rounded off)

Hence required percentage = 0.38x100 = 38%

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Richard and Teo have a combined age of 37. Richard is 4 years older than twice​ Teo's age. How old are Richard and​ Teo?
valkas [14]

Answer:

Teo is 11 years old and Richard is 26 years old.

Step-by-step explanation:

Let R be the age of Richard and T be the age of Teo.

<u><em>First set up the equations:</em></u>

The combined age of Teo and Richard is 37:

R+T=37

Richard is four years old than twice Teo's age:

R=2T+4

<u><em>Then, solve the equations:</em></u>

Substitute (2) to (1) and solve for T:

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5 0
3 years ago
Please help on this one?
AleksandrR [38]

The choices you provided do not match the question!!!

Tickets ($25): x

T-shirts ($12): y

25x + 12y + 4 ≤ 130

Since she bought 3 tickets: 25(3) + 12y + 4 ≤ 130

75 + 12y + 4 ≤ 130

79 + 12y ≤ 130

12y ≤ 51

y ≤ \frac{51}{12}

y ≤ 4\frac{1}{4}

Since you cant buy \frac{1}{4} of a T-shirt, the maximum number of T-shirts she can purchase is 4.


3 0
3 years ago
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