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tatyana61 [14]
4 years ago
13

In the past, 35% of the students at ABC University were in the Business College, 35% of the students were in the Liberal Arts Co

llege, and 30% of the students were in the Education College. To see whether or not the proportions have changed, a sample of 300 students from the university was taken. Ninety of the sample students are in the Business College, 120 are in the Liberal Arts College, and 90 are in the Education College. Using α = .05, the conclusion of the test is that the a. proportions have not changed significantly. b. proportions follow normal distribution. c. Marascuilo procedure is more applicable. d. null hypothesis cannot be rejected.
Mathematics
1 answer:
sukhopar [10]4 years ago
8 0

Answer:

a. proportions have not changed significantly

Step-by-step explanation:

Given

Business College= 35 %

Arts College= 35 %

Education College = 30%

Calculated

Business College = 90/300= 9/30= 0.3 or 30%

Arts College= 120/300= 12/30= 2/5= 0.4 or 40%

Education College= 90/300= 9/30 = 0.3 or 30%

First we find the mean and variance of the three colleges using the formulas :

Mean = np

Standard Deviation= s= \sqrt{npq\\}

Business College

Mean = np =300*0.3= 90

Standard Deviation= s= \sqrt{npq\\}=\sqrt{0.3*0.7*300}= 7.94

Arts College

Mean = np =300*0.4= 120

Standard Deviation= s= \sqrt{npq\\}=\sqrt{0.4*0.6*300}=  8.49

Education College

Mean = np =300*0.3= 90

Standard Deviation= s= \sqrt{npq\\}=\sqrt{0.3*0.7*300}= 7.94

Now calculating the previous means with the same number of students

Business College

Mean = np =300*0.35= 105

Arts College

Mean = np =300*0.35= 105

Education College:

Mean = np =300*0.3= 90

Now formulate the null and alternative hypothesis

Business College

90≤ Mean≥105

Arts College

105 ≤ Mean≥ 120

Education College

U0 : mean= 90     U1: mean ≠ 90

From these we conclude that the  proportions have not changed significantly meaning that it falls outside the critical region.

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The school that Michael goes to is selling tickets to a dance. On the first day of ticket sales the school sold 3 staff tickets
Minchanka [31]

The price of a staff ticket and the price of a student ticket is $8 and $14

Given:

Day 1:

Number of staff tickets sold = 3

Number of students tickets sold = 1

Total revenue day 1 = $38

Day 2:

<em>Number of staff tickets sold</em> = 3

<em>Number of students tickets sold</em> = 2

<em>Total revenue day</em> 2 = $52

let

<em>cost of staff tickets</em> = x

<em>cost of students tickets</em> = y

The equation:

<em>3x + y = 38 (1)</em>

<em>3x + y = 38 (1)3x + 2y = 52 (2)</em>

subtract (1) from (2)

2y - y = 52 - 38

y = 14

substitute y = 14 into (1)

3x + y = 38 (1)

3x + 14 = 38

3x = 38 - 14

3x = 24

x = 24/3

x = 8

Therefore,

cost of staff tickets = x

= $8

cost of students tickets = y

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8 0
3 years ago
If you only have a
cestrela7 [59]

Hi! I'm happy to help!

To solve this problem, we need to divide the recipe amount in 1/6 amounts. So, we will do a fraction division problem like this:

15\frac{1}{6}÷\frac{1}{6}

This problem is hard to do with mixed numbers, so we need to turn 15\frac{1}{6} into an improper fraction. To do that we need to multiply 15 by 6, because that is our denominator, then add the extra \frac{1}{6}.

(15×6)+1

90+1

91

So, our improper fraction would be\frac{91}{6}, now, let's solve.

\frac{91}{6}÷\frac{1}{6}

It is difficult to do division problems on their own, so we can change this into an easier problem. We can do the inverse operation and turn this into multiplication. We do this by changing it to multiplication (obviously), then flip the second fraction.

\frac{91}{6}×\frac{6}{1}

Now, we just multiply the top by the top, and bottom by the bottom.

\frac{546}{6}

We could end it here, but we want a whole number, so, we simplify the number by dividing both the top and bottom by 6.

\frac{91}{1}

Anything over 1, is just a whole number

91.

<u>Therefore, the recipe should require 91 uses of the 1/6 cup.</u>

I hope this was helpful, keep learning! :D

3 0
3 years ago
What is the sum or difference?<br> a. 5x4 + 4x4<br> b. 12x5y − 9x5y
nydimaria [60]

Answer:

a. 9x^4

b. 3 x^5y

Step-by-step explanation:

a. 5x^4 + 4x^4

Combine like terms

9x^4

b. 12x^5 y − 9x^5 y

Subtract like terms

  3 x^5y

4 0
3 years ago
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