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Dovator [93]
3 years ago
8

A standardized test was given to a set of high school juniors and the distribution of the data is bell shaped. The mean score is

800 and the standard deviation is 120.
To qualify for a special summer camp for accelerated students, a student must score within the top 16% of all scores on the test. What score must a student make to qualify for summer camp?
Mathematics
1 answer:
stira [4]3 years ago
5 0

Answer:

920 points.      

Step-by-step explanation:    

We have been given that the mean score for a standardized test is 800 and the standard deviation is 120. To qualify for a special summer camp for accelerated students, a student must score within the top 16% of all scores on the test.

First of all we will find probability of 0.16 using normal distribution table.

Using normal distribution our Z score will be 0.994458

Now we will use raw-score formula to find the score (x) that a student must make to qualify for summer camp.  

x=\text{Mean}+\text{ Standard deviation* Z score}

Upon substituting our given values in above formula we will get,

x=800+120\times 0.994458

x=800+119.33496

x=919.33496

Upon rounding to nearest whole number we will get,

x=920

Therefore, a student must make 920 points to qualify for summer camp.

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Find the least common multiple of x2 – 4x – 5 and x2 – 3x – 10.
vfiekz [6]

<u>x² - 4x - 5</u>                           <u>x² - 3x - 10</u>

(x - 5)(x + 1)                         (x - 5)(x + 2)

                  GCF: (x - 5)

                  LCM: (x - 5)(x + 1)(x + 2)

Answer: (x - 5)(x + 1)(x + 2)

5 0
3 years ago
A movie theater has a seating capacity of 329. The theater charges $5.00 for children, $7.00 for students, and $12.00 of adults.
kipiarov [429]

Answer:

170 children

74 students

85 adults

Step-by-step explanation:

Given

Let:

C = Children; S = Students; A = Adults

For the capacity, we have:

C + S + A = 329

For the tickets sold, we have:

5C + 7S + 12A = 2388

Half as many as adults are children implies that:

A = \frac{C}{2}

Required

Solve for A, C and S

The equations to solve are:

C + S + A = 329 -- (1)

5C + 7S + 12A = 2388 -- (2)

A = \frac{C}{2} -- (3)

Make C the subject in (3)

C = 2A

Substitute C = 2A in (1) and (2)

C + S + A = 329 -- (1)

2A + S + A = 329

3A + S = 329

Make S the subject

S = 329 - 3A

5C + 7S + 12A = 2388 -- (2)

5*2A + 7S + 12A = 2388

10A + 7S + 12A = 2388

7S + 22A = 2388

Substitute S = 329 - 3A

7(329 - 3A) + 22A = 2388

2303 - 21A + 22A = 2388

2303 +A = 2388

Solve for A

A = 2388 - 2303

A = 85

Recall that: C = 2A

C = 2 * 85

C = 170

Recall that: S = 329 - 3A

S = 329 - 3 * 85

S = 329 - 255

S = 74

Hence, the result is:

C = 170

S = 74

A = 85

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