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Evgesh-ka
1 month ago
10

1. Hank is an intelligent student and usually makes good grades, provided that he can review the course material the night befor

e the test. For tomorrow's test, Hank is faced with a small problem: His fraternity brothers are having an all-night party in which he would like to participate. Hank has three options: a1 - party all night: 2-divide the night equally between studying and partying: 3 - study all night. Tomorrow's exam can be easy (s1), moderate (S2) or tough (53), depending on the professor's unpredictable mood. Hank anticipates the following scores:
S1 S2 S3
a1 85 60 40
a2 92 85 81
a3 100 88 82

(a) Recommend a course of action for Hank based on each of the four criteria of decisions under uncertainty.
(b) Suppose that Hank is more interested in the letter grade he will get. The dividing scores for the passing letter grades A to Dare 90, 80, 70 and 60, respectively. What should the decision/s be?

Mathematics
1 answer:
PIT_PIT [12.4K]1 month ago
3 0

Answer and Detailed Explanation:

Below is the response provided

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I JUST NEED THESE 3 QUESTIONS
AnnZ [12381]

Answer:  

 

1.     What are the amplitude and period of the sine function that indicates the positioning of the band members as they start performing?

Answer:  The amplitude is 80 ft and the period is 60 ft.

 

2.     Edna, seated in the stands, faces Darla and notices that the sine curve starts rising from the left edge of the field. What is the equation for the sine function representing the arrangement of band members at the beginning of their performance?

Answer:  y = 80cos(x*π/30)+80

 

3.     When the band starts playing, the members move away from the edges, and the sine curve changes to start decreasing at the far left. Darla remains in her position. Now the sine curve is half as tall as it originally was. What is the equation for the updated sine curve depicting the band members' positions?

Answer: y = 40cos(x*π/30)+80

 

4.     Finally, the entire band shifts closer to the edge of the football field, causing the sine curve to now position itself in the lower half of the field from Edna’s perspective. What is the equation for this sine curve reflecting the band members' positions after these adjustments?

Answer:  y = 40cos(x*π/30)+40

Step-by-step explanation:

8 0
1 month ago
A certain pen has been designed so that true average writing lifetime under controlled conditions (involving the use of awriting
tester [12383]

Answer:

a) Null hypothesis:\mu \geq 10

Alternative hypothesis:\mu < 10

b) p_v =P(t_{17}

Given that the p-value is lower than the significance threshold in this situation, we have sufficient grounds to reject the null hypothesis.

c) p_v =P(t_{17}

In this case, since the p-value exceeds the significance threshold, we have adequate evidence to FAIL to reject the null hypothesis.

d) p_v =P(t_{17}

Here again, with the p-value being less than the significance level, we can reject the null hypothesis.

Step-by-step explanation:

1) Provided data and references

\bar X represents the average of the samples

s denotes the standard deviation of the samples

n=18 indicates the number of samples

\mu_o =10 is the value we are examining

\alpha defines the significance level for the test.

t represents the specific statistic of interest

p_v indicates the p-value relevant to the test (the variable of concern)

Define the null and alternative hypotheses.

To assess if the true mean is at least 10 hours, we must set up a hypothesis:

Part a

Null hypothesis:\mu \geq 10

Alternative hypothesis:\mu < 10

If we consider the sample size being less than 30 and the population deviation unknown, it’s more appropriate to use a t-test to compare the actual mean with the reference value, calculated as:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)

Part b

In this scenario t=-2.3, \alpha=0.05

Initially, we need to calculate the degrees of freedom df=n-1=18-1=17

Since this is a left-tailed test, the p-value is determined by:

p_v =P(t_{17}

In this instance, with the p-value being less than the significance level, we have sufficient evidence to reject the null hypothesis.

Part c

For this situation t=-1.8, \alpha=0.01

We need to find the degrees of freedom df=n-1=18-1=17

For the left-tailed test, the p-value is given by:

p_v =P(t_{17}

In this case, since the p-value is above the significance level, we have enough grounds to FAIL to reject the null hypothesis.

Part d

For this case t=-3.6, \alpha=0.05

Firstly, we find the degrees of freedom df=n-1=18-1=17

Since we are conducting a left-tailed test, the p-value is calculated as:

p_v =P(t_{17}

Here, with the p-value being lower than the significance threshold, we can reject the null hypothesis.

5 0
1 month ago
Decimal 0.003 is 1/10 of
zzz [12365]
To determine the result of multiplying 0.003 by 10, you will simply multiply as such. A straightforward method is to shift the decimal point one place to the right (it would shift two places for multiplying by 100, and three for 1000, etc.). The same rule applies when dividing; just ensure to move left instead.

Hence, the result is 0.03.

<span>I hope that was helpful!!!</span>
6 0
1 month ago
Sara’s little brother is getting ready to color the picture of a rainbow that Sara drew for him. To make the task easier for her
Zina [12379]
There are seven rainbow colors: red, orange, yellow, green, blue, indigo, and violet, so 7 possible choices. When two events occur in sequence, multiply their probabilities. With replacement: P(violet)=1/7 and P(orange)=1/7, so P(violet then orange)=1/7 * 1/7 = 1/49. Without replacement: after picking violet, P(orange)=1/6, so P(violet then orange)=1/7 * 1/6 = 1/42.
8 0
15 days ago
Read 2 more answers
in the drawing six out of every 10 tickets are winning tickets of the winning tickets one out of every three awards is a larger
tester [12383]

Answer: P(\text{a ticket is randomly chosen will award a large prize)}=\frac{2}{21}

Given that

Total tickets = 10

Winning tickets = 6

\text {Possibility of getting a winning ticket }=^{10}C_6

For every three prizes, there’s one large prize

therefore,

\text{ Possibility of getting a large prize }=^6C_3

At this point,

P(\text{a ticket is randomly chosen will award a large prize)}=\frac{^6C_3}{^{10}C_6}\\\\=\frac{20}{210}\\\\=\frac{2}{21}

P(\text{a ticket is randomly chosen will award a large prize)}=\frac{2}{21}

8 0
20 days ago
Read 2 more answers
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