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KatRina [158]
2 years ago
12

Lisa bought a tank for her hermit crab. The tank can hold 1,331 cubic inches of water and includes a top. The base of the aquari

um is a square. The height of the aquarium is 11 inches. A: what is the length of each side of the base of the tank. B: what shape is Lisa’s tank. D what is the surface area of the tank
Mathematics
1 answer:
Alina [70]2 years ago
5 0

Answer:

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Step-by-step explanation:

<h2>event type motion detection event time you are you </h2>
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F(x) = 2x + 5 then f(?) = 5<br><br> 2<br> 1<br> 0<br> 3
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Answer: answer is 0

Step-by-step explanation:

F(0)=2(0)+5

F(0)=0+5

F(0)=5

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At a meeting, there were 18 men and 24 women. What is the ratio of men to women at the meeting?
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men = 12

women = 8

Step-by-step explanation:

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A recreation center charges nonmembers $3 to use the pool and $5 to use the basketball courts. A person pays $42 to use the recr
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Let x = times in the pool
Let y = times in the basketball courts

42=3x+5y
x+y=12
Therefore y = 12-x
Subsitute for y
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42 = 3x + 60 -5x
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-18 = -2x
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The person used the pool 9 times.

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4.One attorney claims that more than 25% of all the lawyers in Boston advertise for their business. A sample of 200 lawyers in B
AleksAgata [21]

Answer:

z=\frac{0.315 -0.25}{\sqrt{\frac{0.25(1-0.25)}{200}}}=2.123  

p_v =P(Z>2.123)=0.0169  

The p value obtained was a very low value and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of lawyers had used some form of advertising for their business is significantly higher than 0.25 or 25% .  

Step-by-step explanation:

1) Data given and notation  

n=200 represent the random sample taken

X=63 represent the lawyers had used some form of advertising for their business

\hat p=\frac{63}{200}=0.315 estimated proportion of lawyers had used some form of advertising for their business

p_o=0.25 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 represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that more than 25% of all the lawyers in Boston advertise for their business:  

Null hypothesis:p\leq 0.25  

Alternative hypothesis:p > 0.25  

When we conduct a proportion test we need to use the z statistic, 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.

3) Calculate the statistic  

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

z=\frac{0.315 -0.25}{\sqrt{\frac{0.25(1-0.25)}{200}}}=2.123  

4) 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=0.05. The next step would be calculate the p value for this test.  

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

p_v =P(Z>2.123)=0.0169  

The p value obtained was a very low value and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of lawyers had used some form of advertising for their business is significantly higher than 0.25 or 25% .  

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3 years ago
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