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2 years ago

What percent of 33 is 34

Mathematics

2 answers:

ivann1987 [24]2 years ago

5 0

I think it is 34 is 103.030303% of 33

Hope this helps :/

Viktor [21]2 years ago

3 0

11.22 is 33% of 34 if that is what youre asking

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A screening test for a newly discovered disease is being evaluated for its performance as a screening test. In order to determin

Elena-2011 [213]

Answer:

- Sensitivity = 62.7 %

- specificity = 74.9 %

- positive predictive value = 45.2 %

- negative predictive value = 85.9 %

- Accuracy of the test = 71.9 %

Step-by-step explanation:

Given the data in th question;

CONDITION POSITIVE CONDITION NEGATIVE TOTAL

TEST POSITIVE 42 51 93

TEST NEGATIVE 25 152 177

TOTAL 67 203 270

so; A=42, B=51, C=25 and D=152

Sensitivity = [A / ( A + C) ]×100= [42 /( 42 + 25)]×100 = [42 / 67] 100=

Sensitivity = 62.7 %

specificity = [D / ( D + B) ]×100= [152 /( 152 + 51)]×100 = [152 / 203] 100=

specificity = 74.9 %

positive predictive value = [A / ( A + B) ]×100 = [42 /( 42 + 51)]×100 =[42/93] 100

positive predictive value = 45.2 %

negative predictive value = [D / ( D + C) ]×100 = [152 /( 152 + 25)]×100 =[152/177] 100

negative predictive value = 85.9 %

Accuracy of the test = [A + D / ( A+B+C+D)] = [42 + 152 / ( 42+51+25+152)]

= [ 194 / 270 ]

Accuracy of the test = 71.9 %

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2 years ago

Sophia injured her leg and had to go to the doctor. She paid the bills with her credit card. What is a disadvantage of using the

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Which expression best represents the opposite of the opposite of 8 1/4

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2 years ago

According to the February 2008 Federal Trade Commission report on consumer fraud and identity theft, 23% of all complaints in 20

Murljashka [212]

Answer:

There is not enough evidence to support the claim that Alaska had a lower proportion of identity theft than 23%.

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 1432

p = 23% = 0.23

Alpha, α = 0.05

Number of theft complaints , x = 321

First, we design the null and the alternate hypothesis

$H_{0}: p = 0.23\\H_A: p < 0.23$

This is a one-tailed test.

Formula:

$\hat{p} = \dfrac{x}{n} = \dfrac{321}{1432} =0.2241$

$z = \dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}$

Putting the values, we get,

$z = \displaystyle\frac{0.2241-0.23}{\sqrt{\frac{0.23(1-0.23)}{1432}}} = -0.5305$

Now, we calculate the p-value from the table.

P-value = 0.298

Since the p-value is greater than the significance level, we fail to reject the null hypothesis and accept the null hypothesis.

Conclusion:

Thus, there is not enough evidence to support the claim that Alaska had a lower proportion of identity theft than 23%.

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2 years ago

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Answer:

The answer is C. An absolute value will never have a negative in it.

Step-by-step explanation:

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