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grin007 [14]
2 years ago
15

The mean score of 15 students was 82.0. (a) Another student joins, and the mean becomes 82.5. What is the new student's score? (

b) Suppose instead that two students join, taking the average from 82 to 83. What could their scores be, if they are restricted to be integer (i.e. whole) numbers? (Give all the possible solutions, under the assumption that the maximum score for an individual student is 100.)
Mathematics
1 answer:
slega [8]2 years ago
4 0

Answer:

a) The new student's score is 90.

b) Their scores are a pair of scores lower than 100 with a mean of 90.5.

Step-by-step explanation:

The mean is the sum of all scores divided by the number of students.

The mean score of 15 students was 82.0

This means that there are 15 students and the sum of the scores is 15*82 = 1230.

(a) Another student joins, and the mean becomes 82.5. What is the new student's score?

Now there are 16 students, the sum of the scores is 1230 + x and the mean is 82.5. So

82.5 = \frac{1230 + x}{16}

1230 + x = 16*82.5

x = 1320 - 1230

x = 90

The new student's score is 90.

(b) Suppose instead that two students join, taking the average from 82 to 83. What could their scores be, if they are restricted to be integer (i.e. whole) numbers?

Now there are 2 new students, so 17 in total. The mean score of the new students is 2x. The total score of the class is 1230 + 2x and the mean is 83. So

83 = \frac{1230 + 2x}{17}

1230 + 2x = 1411

2x = 181

x = 90.5

The mean of the two students scores is 90.5.

So their scores are a pair of scores lower than 100 with a mean of 90.5.

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A probability calculator is required on this problem; answer to six decimal places. Suppose we will spin the wheel pictured 400
KiRa [710]

Answer:

P(90< X< 110)= P(\frac{90-80}{8}

And we can find this probability with this difference:

P(90< X< 110)=P(z

And we can find the real value with the following excel code using the binomial distribution:

"=BINOM.DIST(110,400,0.2,TRUE)-BINOM.DIST(89,400,0.2,TRUE)"

And we got 0.118 a very close value from the value obtained using the normal approximation

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Let X the random variable of interest, on this case we now that:

X \sim Binom(n=400, p=0.2)

The probability mass function for the Binomial distribution is given as:

P(X)=(nCx)(p)^x (1-p)^{n-x}

Where (nCx) means combinatory and it's given by this formula:

nCx=\frac{n!}{(n-x)! x!}

We need to check the conditions in order to use the normal approximation.

np=400*0.2=80 \geq 10

n(1-p)=400*(1-0.2)=320 \geq 10

So we see that we satisfy the conditions and then we can apply the approximation.

If we appply the approximation the new mean and standard deviation are:

E(X)=np=400*0.2=80

\sigma=\sqrt{np(1-p)}=\sqrt{400*0.2(1-0.2)}=8

So then we can approximate the random variable as X \sim N(\mu = 80, \sigma = 8)

And we want this probability:

P(90< X< 110)

We can use the z score formula given by:

z = \frac{x -\mu}{\sigma}

And replacing we got:

P(90< X< 110)= P(\frac{90-80}{8}

And we can find this probability with this difference:

P(90< X< 110)=P(z

And we can find the real value with the following excel code using the binomial distribution:

"=BINOM.DIST(110,400,0.2,TRUE)-BINOM.DIST(89,400,0.2,TRUE)"

And we got 0.118 a very close value from the value obtained using the normal approximation

8 0
2 years ago
A coach chooses six out of eight players to go to a skills workshop. if order does not matter, in how many ways can he choose th
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2 years ago
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Disney held a breakfast for parents and their children to eat with Mickey mouse. adult tickets cost $17.95 and children's ticket
RUDIKE [14]

Answer:

The number of children attending = 324

The number of adults attending  = 176.

Step-by-step explanation:

Here, the total number of people attending = 500

Let us assume the umber of children attending the breakfast = m

So, the number of adults attending the breakfast = 500 - m

Cost of ticket for each children  = $12.95

So, the cost of m children's tickets = m x ( Cost of 1 children ticket)

=  m x ( $12.95)  = 12.95 m .... (1)

Cost of ticket for each adult  = $17.95

So, the cost of (500 -m)  adults's tickets

= (500 - m) x ( Cost of 1 adult ticket)    = (500 -m) x ( $17.95)  

 = 8,975 - 17.95 m     .... (2)

Now, the total earnings from the total ticket sold = $7355

So, The earning from ( Adult's + children's tickets)  = $7355

⇒  12.95 m  +  8,975 - 17.95 m = $7355

or, - 5 m = 7355 - 8975 =   -1620

or, m = 1620/5 = 324

⇒  m =324

Hence the number of children attending =  m = 324

The number of adults attending = 500 - m = 500 - 324 = 176.

7 0
2 years ago
Suppose you have just enough money, in coins, to pay for a gallon of milk priced at $2.40. You have 12 coins, all quarters and d
wel
The answer is
q + d = 12
0.25q + 0.10d = 2.40


q - the number of quarters
d - the number of dimes

You have 12 <span>coins, all quarters and dimes:
q + d = 12

The value of 1 quarter is $0.25.
The value of 1 dime is $0.10.
Therefore, if you want to pay </span>$2.40 you will have q quarters of value $0.25(0.25q) and d dimes of value $0.10d (0.10d).
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2 years ago
1. Darnell reads at a constant rate
JulsSmile [24]

Answer:

Darnell can read 1,715 words in 7 minutes

Step by step Explanation:

1. I need to find out how many words he reads in 1 minute. Divide 735 by 3, my answer is 245

245

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3)735

6 bring down the 3 to make 13

-_____

1 3

12 bring down the 5 to make 15

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1 5

15

___________

0

2. multiply 735 by 2 cause he reads 735 words in 3 minutes and we are tryna find out how many words he reads in 7 minutes. 3×2=6, so 735+735 (735×2) equals 1470

3. Add 245, for that extra minute that we missed. my answer is 1,715 words in 7 minutes

8 0
2 years ago
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