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andreev551 [17]
3 years ago
10

david took 10 1/2 hours over 3 days to build airplane. he spent 3 1/4 hrs the firsr day and 4 2/3 hours the second day. how many

hours did he spend the third day to finish?
Mathematics
1 answer:
Anuta_ua [19.1K]3 years ago
8 0
2 7/12 hours

X=101/2 - ( 3+4+1/4+2/3)

X=101/2 - 7 + (1/4+2/3)

1/4+2/3= 11/12

101/2 - 7 =3 1/5

X = 3 1/5 - 11/12 = 2 7/12
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Step-by-step explanation:

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I'm pretty sure its line c.

Step-by-step explanation:

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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.

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2 years ago
Chester Boles reviewed his credit card's monthly statement. It shows a $2,376.10 previous balance and this month's new purchases
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Initial balance, I = $2376.10 .

Total amount of purchase made, A = $( 875.22+65.75+45.22+21.23 ) = $1007.42 .

Total amount credit, c = $875.22 .

Fine, f = $45.30 .

Another purchase, a=\dfrac{2376.10\times 2.5}{100}=\$ 59.4025 .

So, balance left is :

B = I - A - f - a + c

B = 2376.10 - 1007.42 - 45.30 - 59.4025 + 875.22

B = $2139.1975

Hence, this is the required solution.

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