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kobusy [5.1K]
3 years ago
13

A theater where a drug abuse program is being presented seats 150 people. The proceeds will be donated to a local drug informati

on center. Admission is $2.00 for adults and $1.00 for students. Every two adults must bring at least one student. How many adults and students should attend in order to raise the maximum amount of money?
Mathematics
1 answer:
Yuliya22 [10]3 years ago
4 0

there should be at least 100 adults and 50 students because if there are 100 hundred adults- that means there would be 1 student per 2 adults giving you 100 adults and 50 student equaling a total of 150 seats which is the max seats the theater can hold and that would raise


$2.00 x 100 = $200.00

$1.00 x 50 = $50.00

$200.00 + $50.00 = $250.00

$250.00 is the max amount of money the program can make and to make that $250.00--- 100 adults and 50 students would have to attend

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The diameters of cherry tomatoes produced by a large farm have an approximately Normal distribution, with a
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Answer: D. 0.9967

Step-by-step explanation: To solve this, you need to do the z-score formula with both numbers. So 30-22/2.5 = 3.2. On the z-score chart, that equals .9993. Hold onto that number. Then you do the same with 15. 15-22/2.5 = -2.8. On the z score chart, that equals .0026. Subtract those numbers. .9993 - .0026 = .9967. It's not as compacted as it looks <3

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Which histogram represents the data in the frequency table showing the age of people leaving the library?
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Answer:

last graph

Step-by-step explanation:

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HELPPPPP
OlgaM077 [116]

Answer:

Point on Midline (0,3)

Maximum (9π/2,3)

Minimum (-9π/2,-3)

Step-by-step explanation:

in the given sine function which is in the form of f(x) = a sin(bx+c) +d

a = amplitude

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Therefore b = 2π/18π = 1/9

Y intercept = vertical shift = 3

Horizontal shift = d = 0

Therefore the sine function will be

f(x) = 6 sin(x/9) + 3

Now first point on the midline is (0,3)

Second point is maximum (9π/2,9)

Third point be a minimum value ( -9π/2,-3)

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Read 2 more answers
Women athletes at the a certain university have a long-term graduation rate of 67%. Over the past several years, a random sample
lana [24]

Answer:

z=\frac{0.579 -0.67}{\sqrt{\frac{0.67(1-0.67)}{38}}}=-1.193  

p_v =P(z  

So the p value obtained was a very high value and using the significance level given \alpha=0.1 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can conclude that  the proportion of women athletes graduated is not significantly lower than 0.67 or 67% at 10% of significance

Step-by-step explanation:

Data given and notation

n=38 represent the random sample taken

X=22 represent the number of women athletes graduated

\hat p=\frac{22}{38}=0.579 estimated proportion of women athletes graduated

p_o=0.67 is the value that we want to test

\alpha=0.1 represent the significance level

Confidence=90% or 0.90

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is lower than 0.67 or no:  

Null hypothesis:p \geq 0.67  

Alternative hypothesis:p < 0.67  

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)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

Calculate the statistic  

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

z=\frac{0.579 -0.67}{\sqrt{\frac{0.67(1-0.67)}{38}}}=-1.193  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided \alpha=0.1. The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

p_v =P(z  

So the p value obtained was a very high value and using the significance level given \alpha=0.1 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can conclude that  the proportion of women athletes graduated is not significantly lower than 0.67 or 67% at 10% of significance

3 0
2 years ago
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