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eimsori [14]
4 years ago
3

Lena is playing her accordion by compressing and stretching it. The accordion's length (in cm) t seconds

Mathematics
1 answer:
mixas84 [53]4 years ago
3 0

Answer:

It takes Lena 4.1 seconds to compress the accordion completely and then stretch it back out to a length  of 50cm.

Step-by-step explanation:

I am assuming the function without a T: A(t)= 25cos(t)+65.

The function is graphed in the figure attached, and the blue line is y =50cm.

From the graph we see that the accordion is completely stretched  (has minimum length) at t= \pi seconds, and stretches back  to 50cm at t=4.069s; therefore, it takes Lena 4.069 seconds to stretch the accordion back again, which to the nearest tenth of a second is  4.1s

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Read 2 more answers
A civil engineer is analyzing the compressive strength of concrete. Compressive strength is normally distributed with A random s
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Answer:

95%: (3278.354 ; 3270.083)

99% : (3221.646 ; 3278.354)

Step-by-step explanation:

Given :

Sample size, n = 12

Mean, xbar = 3250

Sample standard deviation = √1000

The 95% confidence interval :

Mean ± Margin of error

Margin of Error = Tcritical * s/√n

Tcritical at 0.05, df=12-1 = 11 ;

Tcritical at 95% = 2.20

Hence,

Margin of Error = (2.20 * √1000/√12) = 20.083

Confidence interval : 3250 ± 20.083

Lower boundary = 3250 - 20.083 = 3229.917

Upper boundary = 3250 + 20.083 = 3270.083

2.)

The 99% confidence interval :

Mean ± Margin of error

Margin of Error = Tcritical * s/√n

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Tcritical at 99% = 3.106

Hence,

Margin of Error = (3.106 * √1000/√12) = 28.354

Confidence interval : 3250 ± 28.354

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Upper boundary = 3250 + 28.354 = 3278.354

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3 years ago
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