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3 years ago

Change 33/12 to a mixed number

Mathematics

2 answers:

LuckyWell [14K]3 years ago

8 0

Answer:

2 9/12

Explanation:

Divide using long division. The whole number portion will be the number of times the denominator of the original fraction divides evenly into the numerator of the original fraction, and the fraction portion of the mixed number will be the remainder of the original fraction division over the denominator of the original fraction.

Hope this helps. :-)

Aleks04 [339]3 years ago

6 0

Answer:

2 3/4

Step-by-step explanation:

To change an improper fraction to a mixed number

Take 33 and divide by 12

12 goes into 33 two times

2*12 =24

33-24 = 9

2 9/12

We can simplify the fraction

Divide the top and bottom by 3

9/3 =3

12/3 =4

The mixed number becomes

2 3/4

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A company makes Citi bikes. 95% (i.e., 0.95) pass final inspection. Suppose that 5 bikes are randomly selected.

Phoenix [80]

Answer:

Hi myself Shrushtee.

Step-by-step explanation:

Given,

p=0.95

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8 0

3 years ago

Please help me it's urgent!

Crank

Answer:

y = 2x-5

y=-x+4

Step-by-step explanation:

Just need to solve the:

x+y=4

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3 years ago

A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than ex

Olenka [21]

Answer:

The 90% confidence interval that estimate the true proportion of students who receive financial aid is

$0.533 < p < 0.64$

$n = 1789$

Step-by-step explanation:

Considering question a

From the question we are told that

The sample size is n = 200

The number of student that receives financial aid is $k = 118$

Generally the sample proportion is

$\^ p = \frac{k}{n}$

=> $\^ p = \frac{118}{200}$

=> $\^ p = 0.59$

From the question we are told the confidence level is 90% , hence the level of significance is

$\alpha = (100 - 90 ) \%$

=> $\alpha = 0.10$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.645$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{\frac{\^ p (1- \^ p)}{n} }$

=> $E = 1.645 * \sqrt{\frac{0.59 (1- 0.59)}{200} }$

=> $E = 0.057$

Generally 90% confidence interval is mathematically represented as

$\^ p -E < p < \^ p +E$

=> $0.533 < p < 0.64$

Considering question b

From the question we are told that

The margin of error is E = 0.03

From the question we are told the confidence level is 99% , hence the level of significance is

$\alpha = (100 - 99 ) \%$

=> $\alpha = 0.01$

Generally from the normal distribution table the critical value of is

$Z_{\frac{\alpha }{2} } = 2.58$

Generally the sample size is mathematically represented as

$[\frac{Z_{\frac{\alpha }{2} }}{E} ]^2 * \^ p (1 - \^ p )$

=> $n = [\frac{2.58}{0.03} ]^2 * 0.59 (1 - 0.59 )$

=> $n = 1789$

8 0

3 years ago