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nevsk [136]
3 years ago
5

Abby practiced for 11/12 of an hour and Lauren practiced for 2/3 of an hour. How long did they practice on Wednesday

Mathematics
2 answers:
Solnce55 [7]3 years ago
6 0
2/3 equals 8/12

11/12 plus 8/12 will give you the improper fraction of 19/12.

19/12 also equals
1 7/12.
Olin [163]3 years ago
5 0
You first have to equalize all of the denominator

2/3 will be equal to :  8/12

Total of their practice time would be :

11/12 + 8/12    = 19/12 hours

Hope this helps
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a test has 20 multiple-choice questions with 5 choices each, followed by 35 true/false questions. if a student guesses on each q
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There are 170 ways the student can answer the test.

Explanation

If there are 20 multiple-choice questions with 5 choices each, the student has 100 choices. The first question has 5 choices to pick from. The second has 5 as well. So does the third. Hopefully now you realize that you have to multiply the number of choices by the number of questions.

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2 years ago
A survey of 76 commercial airline flights of under 2 hours resulted in a sample average late time for a flight of 2.55 minutes.
nika2105 [10]

Answer:

The best point of estimate for the true mean is:

\hat \mu = \bar X = 2.55

2.55-1.96\frac{12}{\sqrt{76}}=-0.148    

2.55+1.96\frac{12}{\sqrt{76}}=5.248    

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

Step-by-step explanation:

Information given

\bar X=2.55 represent the sample mean for the late time for a flight

\mu population mean

\sigma=12 represent the population deviation

n=76 represent the sample size  

Confidence interval

The best point of estimate for the true mean is:

\hat \mu = \bar X = 2.55

The confidence interval for the true mean is given by:

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}   (1)

The Confidence level given is 0.95 or 95%, th significance would be \alpha=0.05 and \alpha/2 =0.025. If we look in the normal distribution a quantile that accumulates 0.025 of the area on each tail we got z_{\alpha/2}=1.96

Replacing we got:

2.55-1.96\frac{12}{\sqrt{76}}=-0.148    

2.55+1.96\frac{12}{\sqrt{76}}=5.248    

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

7 0
2 years ago
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