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just olya [345]
3 years ago
7

✓ Express 9 inches of rain in 72 hours as a unit rate. *

Mathematics
1 answer:
aksik [14]3 years ago
5 0

Answer: 8

Step-by-step explanation:

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3x-6y=30 solve for x
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Step-by-step explanation:

x = 10+2y

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Darla has three buckets shaped like cylinders. She is filling the buckets
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9 scoops is the correct ans 100% sure

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8. The ratio of the length of Sally's drive to Job A to the length of Sally's
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One number is 32 less than another number. If the sum of the two numbers is 92​, find the two numbers
Aleks04 [339]
<h3>Assumption:</h3>

let the number be x, so that the two numbers are x and x - 32

<h3>Solution:</h3>

x + x - 32 = 92

or, 2x = 92 + 32

or, 2x = 124

or, x = 124/2 = 62

<h2>Answer:</h2>

The first number is x = <u>62.</u>

The second number is x - 32 = 62 - 32 = <u>30</u><u>.</u>

4 0
3 years ago
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It has been reported that the probability that an individual will develop schizophrenia over their lifetime is 0.004. In a rando
Alinara [238K]

Answer:

Null hypothesis:p=0.004  

Alternative hypothesis:p \neq 0.004  

z=\frac{0.00567 -0.004}{\sqrt{\frac{0.004(1-0.004)}{3000}}}=1.449  

p_v =2*P(z>1.149)=0.2506

So the p value obtained was a very high value and using the significance level given \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion if individuals that developed schizophrenia its not significantly different from 0.004.  

Step-by-step explanation:

1) Data given and notation

n=3000 represent the random sample taken

X=17 represent the individuals developed schizophrenia in the sample

\hat p=\frac{17}{3000}=0.00567 estimated proportion of individuals developed schizophrenia in the sample

p_o=0.004 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 true proportion of people who will develop schizophrenia is different from 0.004:  

Null hypothesis:p=0.004  

Alternative hypothesis:p \neq 0.004  

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{0.00567 -0.004}{\sqrt{\frac{0.004(1-0.004)}{3000}}}=1.449  

4) P value

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 bilateral test the p value would be:  

p_v =2*P(z>1.149)=0.2506

5) Decision

So the p value obtained was a very high value and using the significance level given \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion if individuals that developed schizophrenia its not significantly different from 0.004.  

4 0
3 years ago
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