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mel-nik [20]
3 years ago
8

Admission to a movie theater is $10 for children and $11 for adults. On a certain day, 43 people bought tickets at the theater a

nd $452 was collected. How many children's tickets were sold and how many adult's tickets were sold?
Mathematics
1 answer:
Sav [38]3 years ago
4 0

Answer:

21 children's tickets were sold and 22 adult's tickets were sold

Step-by-step explanation:

Create a system of equations where c is the number of children's tickets sold and a is the number of adult's tickets sold:

10c + 11a = 452

c + a = 43

Solve by elimination by multiplying the bottom equation by -10:

10c + 11a = 452

-10c - 10a = -430

Add them together and solve for a:

-a = -22

a = 22

So, 22 adult's tickets were sold. Plug this into one of the equations and solve for c:

c + a = 43

c + 22 = 43

c = 21

So, 21 children's tickets were sold and 22 adult's tickets were sold

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A line that passes through the origin also passes through the point (6,2). What is the slope of the line?
vagabundo [1.1K]

Answer:

slope=1/3

Step-by-step explanation:

y=mx+b

2=m*6+0

m=1/3

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3 years ago
PLEASE HELP (image included) Graph the solution of the equation b - 11 = -8
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It is the third number line.

b-11=-8

What minus 11 would equal -8? 3!

8+3=11 so it would make sense that the dot would be on 3.

smaller numbers minus larger numbers (such as this equation) always end up being negative, so if you have the answer and at least one of the other numbers, it will be easy to figure out.

In this case you would want to maybe just imagine 11-8=? or go 8+?=11.

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3 years ago
Based on the Nielsen ratings, the local CBS affiliate claims its 11:00 PM newscast reaches 41 % of the viewing audience in the a
ZanzabumX [31]

Answer:

1) Null hypothesis:p\geq 0.41  

Alternative hypothesis:p < 0.41

2) \hat p=0.36 estimated proportion of people indicated that they watch the late evening news on this local CBS station

3) z_{crit}=-2.33

And we can use the following excel code to find it: "=NORM.INV(0.01,0,1)"

4) z=\frac{0.36 -0.41}{\sqrt{\frac{0.41(1-0.41)}{1000}}}=-1.017  

5) z_{crit}=-1.28

And we can use the following excel code to find it: "=NORM.INV(0.1,0,1)"

6) We see that |t_{calculated}| so then we have enough evidence to FAIL to reject the null hypothesis at 1% of significance.

7) Null hypothesis:p\geq 0.41  

Step-by-step explanation:

Data given and notation  

n=100 represent the random sample taken

X represent the people indicated that they watch the late evening news on this local CBS station

\hat p=0.36 estimated proportion of people indicated that they watch the late evening news on this local CBS station

p_o=0.41 is the value that we want to test

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

p_v{/tex} represent the p value (variable of interest)  Part 1We need to conduct a hypothesis in order to test the claim that 11:00 PM newscast reaches 41 % of the viewing audience in the area:  Null hypothesis:[tex]p\geq 0.41  

Alternative hypothesis:p < 0.41

Part 2  

\hat p=0.36 estimated proportion of people indicated that they watch the late evening news on this local CBS station

Part 3

Since we have a left tailed test we need to see in the normal standard distribution a value that accumulates 0.01 of the area on the left and on this case this value is :

z_{crit}=-2.33

And we can use the following excel code to find it: "=NORM.INV(0.01,0,1)"

Part 4

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.36 -0.41}{\sqrt{\frac{0.41(1-0.41)}{1000}}}=-1.017  

Part 5

Since we have a left tailed test we need to see in the normal standard distribution a value that accumulates 0.1 of the area on the left and on this case this value is :

z_{crit}=-1.28

And we can use the following excel code to find it: "=NORM.INV(0.1,0,1)"

Part 6

We see that |t_{calculated}| so then we have enough evidence to FAIL to reject the null hypothesis at 1% of significance.

Part 7

Null hypothesis:p\geq 0.41  

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