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alexira [117]
3 years ago
14

Assume that advanced students average 93% on an achievement test and regular students averaged 75%. If 100 advanced students and

300 regular students took the test, what would you expect the average to be? How many of each kind of student would be needed to get a group of 90 students who average 87% on the test? Please include work/explaination, I along with a few of my classmates are confused by this problem. We are in a chapter about systems of equations.
Mathematics
1 answer:
Lemur [1.5K]3 years ago
7 0
Average (mean) = (sum of all the data) / (# of data)

sum of all the data = (average)(# of data)

Thus for 100 students with an average of 93,
sum of all data = (93)(100) = 9300

and for 300 students with an average of 75,
sum of all data = (75)(300) = 22500

Therefore you would expect the overall average to be
(9300 + 22500) / (100 + 300) = 79.5 %

Now if there are x # of advanced students and y # of regular students, then

x + y = 90 (total # of students) and 93x + 75y = 87(x + y) (overall average)

The second equation can be simplified to x - 2y = 0

Subtracting the two equations yields

x = 60 and y = 90

Therefore you would need 60 advanced and 30 regular students.
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The graph below shows the number of open blossoms in a nursery and the number of hours after dawn. answer all that apply
uysha [10]

Answer:

The correct option is B.

Step-by-step explanation:

The given equation is

y=20x+3

This equation represent the number of open blossoms in a nursery after x hours.

At initial the number hours is 0.

Substitute x=0 in the given equation.

y=20(0)+3

y=0+3

y=3

The initial value is 3, it means there were three blossoms open at dawn. Option B is correct.

6 0
3 years ago
A university found that 20% of its students withdraw without completing the introductory statistics course. Assume that 20 stude
EleoNora [17]

Answer:

a) P(X \leq 2)= P(X=0)+P(X=1)+P(X=2)

And we can use the probability mass function and we got:

P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115  

P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576  

P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369  

And adding we got:

P(X \leq 2)=0.0115+0.0576+0.1369 = 0.2061

b) P(X=4)=(20C4)(0.2)^4 (1-0.2)^{20-4}=0.2182  

c) P(X>3) = 1-P(X \leq 3) = 1- [P(X=0)+P(X=1)+P(X=2)+P(X=3)]

P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115  

P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576  

P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369

P(X=3)=(20C3)(0.2)^3 (1-0.2)^{20-3}=0.2054

And replacing we got:

P(X>3) = 1-[0.0115+0.0576+0.1369+0.2054]= 1-0.4114= 0.5886

d) E(X) = 20*0.2= 4

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".  

Solution to the problem  

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

X \sim Binom(n=20, 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!}  

Part a

We want this probability:

P(X \leq 2)= P(X=0)+P(X=1)+P(X=2)

And we can use the probability mass function and we got:

P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115  

P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576  

P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369  

And adding we got:

P(X \leq 2)=0.0115+0.0576+0.1369 = 0.2061

Part b

We want this probability:

P(X=4)

And using the probability mass function we got:

P(X=4)=(20C4)(0.2)^4 (1-0.2)^{20-4}=0.2182  

Part c

We want this probability:

P(X>3)

We can use the complement rule and we got:

P(X>3) = 1-P(X \leq 3) = 1- [P(X=0)+P(X=1)+P(X=2)+P(X=3)]

P(X=0)=(20C0)(0.2)^0 (1-0.2)^{20-0}=0.0115  

P(X=1)=(20C1)(0.2)^1 (1-0.2)^{20-1}=0.0576  

P(X=2)=(20C2)(0.2)^2 (1-0.2)^{20-2}=0.1369

P(X=3)=(20C3)(0.2)^3 (1-0.2)^{20-3}=0.2054

And replacing we got:

P(X>3) = 1-[0.0115+0.0576+0.1369+0.2054]= 1-0.4114= 0.5886

Part d

The expected value is given by:

E(X) = np

And replacing we got:

E(X) = 20*0.2= 4

3 0
3 years ago
Simplify completely quantity x squared plus 4 x minus 45 all over x squared plus 10 x plus 9 and find the restrictions on the va
Brums [2.3K]

Answer:

A) Quantity x minus 5 over quantity x plus 1, where x≠-1 and x≠-9

Step-by-step explanation:

\frac{x^2 + 4x - 45}{x^2 + 10x + 9}

Simplifying the numerator first:

x² + 4x - 45 using the quadratic formula you get;

(x - 5)(x + 9)

Then simplifying the denominator x² + 10x + 9 using a quadratic formula you get;

(x + 1)(x + 9)

Dividing the numerator and denominator now gives;

\frac{(x - 5)(x + 9)}{(x + 1)(x + 9)}

Cancelling (x + 9) throughout leaves you with;

\frac{x - 5}{x + 1}

The only restrictions here is if x = 1 and 9 which will give an undefined answer.

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