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madreJ [45]
3 years ago
7

Find the length of the longest object that will fit inside a cylinder that has a radius of 3.25 inches and is 7 inches high.

Mathematics
1 answer:
ra1l [238]3 years ago
6 0

Answer:

9.5 inches

Step-by-step explanation:

The Pythagorean theorem can be used to find the length of the longest linear object of small dimensions that will fit across the space diagonal of the cylinder. The relevant length is that of the hypotenuse of a triangle whose legs are the diameter and height of the cylinder.

... length² = (3.25×2)² +7² = 91.25

... length = √91.25 ≈ 9.55249 ≈ 9.5 . . . inches

_____

Ordinarily, such a number would be rounded up, but that longer length will not fit. Hence it is appropriate to round the number down.

If the object can be curled up, more than 3.2 billion miles of DNA molecule would fit in the space.

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. (0.5 point) We simulate the operations of a call center that opens from 8am to 6pm for 20 days. The daily average call waiting
SashulF [63]

Answer:

The 95% t-confidence interval for the difference in mean is approximately (-2.61, 1.16), therefore, there is not enough statistical evidence to show that there is a change in waiting time, therefore;

The change in the call waiting time is not statistically significant

Step-by-step explanation:

The given call waiting times are;

24.16, 20.17, 14.60, 19.79, 20.02, 14.60, 21.84, 21.45, 16.23, 19.60, 17.64, 16.53, 17.93, 22.81, 18.05, 16.36, 15.16, 19.24, 18.84, 20.77

19.81, 18.39, 24.34, 22.63, 20.20, 23.35, 16.21, 21.73, 17.18, 18.98, 19.35, 18.41, 20.57, 13.00, 17.25, 21.32, 23.29, 22.09, 12.88, 19.27

From the data we have;

The mean waiting time before the downsize, \overline x_1 = 18.7895

The mean waiting time before the downsize, s₁ = 2.705152

The sample size for the before the downsize, n₁ = 20

The mean waiting time after the downsize, \overline x_2 = 19.5125

The mean waiting time after the downsize, s₂ = 3.155945

The sample size for the after the downsize, n₂ = 20

The degrees of freedom, df = n₁ + n₂ - 2 = 20 + 20  - 2 = 38

df = 38

At 95% significance level, using a graphing calculator, we have; t_{\alpha /2} = ±2.026192

The t-confidence interval is given as follows;

\left (\bar{x}_{1}- \bar{x}_{2}  \right )\pm t_{\alpha /2}\sqrt{\dfrac{s_{1}^{2}}{n_{1}}+\dfrac{s_{2}^{2}}{n_{2}}}

Therefore;

\left (18.7895- 19.5152 \right )\pm 2.026192 \times \sqrt{\dfrac{2.705152^{2}}{20}+\dfrac{3.155945^2}{20}}

(18.7895 - 19.5125) - 2.026192*(2.705152²/20 + 3.155945²/20)^(0.5)

The 95% CI = -2.6063 < μ₂ - μ₁ < 1.16025996668

By approximation, we have;

The 95% CI = -2.61 < μ₂ - μ₁ < 1.16

Given that the 95% confidence interval ranges from a positive to a negative value, we are 95% sure that the confidence interval includes '0', therefore, there is sufficient evidence that there is no difference between the two means, and the change in call waiting time is not statistically significant.

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3 years ago
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Read 2 more answers
Fernando has some nickels, dimes, and quarters that are worth a total of $3.25. He has twice as many dimes as nickels. He has 3
CaHeK987 [17]

Answer:

Number of nickels = 5

Number of dimes = 10

Number of quarters = 8

Step-by-step explanation:

Fernando has a total of nickels + dimes + quarters = 3.25$

Let the number of nickels = x

Then x(.05) + dimes(.10) + quarters(0.25) = 3.25------(1)

He has twice as many dimes as nickels

dimes = 2×nickels

dimes = 2x--------(2)

He has 3 more quarters than nickels

quarters = nickels +3

quarters = x +3------(3)

From equations 1, 2, and 3.

x(.05) + 2x(.10) + (x + 3)(.25) = 3.25

.05x + 0.2x + 0.25x + .75 = 3.25

5x + 20x + 25x + 75 = 325

50x = 325 - 75

50x = 250

x = 5

Therefore the number of nickels are 5

from equation 2 number of dimes are 10

and from equation 3 number of quarters are 8.

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3 years ago
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