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murzikaleks [220]
3 years ago
12

Timothy built a base for a circular tabletop. The base can support a tabletop with a radius of at least 6 inches, but not more t

han 23 inches.
What is the largest possible area of the tabletop that will fit on Timothy’s table base? Round the answer to the nearest whole square inch.
Mathematics
1 answer:
Liono4ka [1.6K]3 years ago
3 0

Answer:

1,662  in²

=============================

Step-by-step explanation:

let the radius of the circular tabletop = r

The base can support a tabletop with a radius of at least 6 inches, but not more than 23 inches.

which mean r at least 6 in. and not more than 23 in.

∴ 6 ≤ r ≤ 23

Since the area of the circle = π * r²

So, the largest possible area of the tabletop will be for r = 23

So, the area = π * 23² = 1,661.9 in²

Round to the nearest whole square inch

So, The largest possible area of the tabletop is 1,662  in²

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A force of 4 pound acts in the direction of 48degree to the horizontal. The force moves an object along a straight line from the
Dafna11 [192]

Answer:

<u>Work Done = 27.66 Joules</u>

Step-by-step explanation:

The work done is:

W=F*d*Cos \theta

Where

F is the fore

d is the distance moved

\theta  is the angle

The force is "4"

The distance is the distance from (1,7) to (8,10)

The distance is the square root of change in y coordinates, squared and sum of change in x coordinates, squared. Hence, distance is:

\sqrt{(10-7)^2+(8-1)^2}\\=\sqrt{3^2+7^2}\\\sqrt{58}

The angle is gotten by the formula:

Cos\phi = \frac{Force in x direction}{Distance}\\Cos\phi = \frac{8-1}{\sqrt{58}}\\Cos\phi=\frac{7}{\sqrt{58}}\\\phi = Cos^{-1}\frac{7}{\sqrt{58}}\\\phi = 23.21

So, now:

\theta = 48 - 23.21 = 24.79

Work Done = F*d * Cos\theta = 4 * \sqrt{58} * Cos(24.79)

<u>Work Done = 27.66 Joules</u>

6 0
3 years ago
The manager of a fleet of automobiles is testing two brands of radial tires and assigns one tire of each brand at random to the
Darya [45]

Answer:

Brand 1    Brand 2    Difference

37734       35202        2532

45299      41635         3664

36240      35500        740

32100      31950         150

37210       38015       −805

48360     47800        560

38200    37810          390

33500    33215        285

Sum of difference = 2532+ 3664+740+150 −805+  560 +390 +285 = 7516

Mean = d=\frac{7516}{8}

Mean = d=939.5

a) d= 939.5

\text{Sample Standard deviation, s} = \sqrt{\dfrac{(x-\bar{x})^2}{n-1}}

=\sqrt{\dfrac{(2532-939.5)^2+(3664-939.5)^2+(740-939.5)^2 ...+(285-939.5)^2}{8-1}}

=1441.21

b)SD= 1441.21

c)Calculate a 99% two-sided confidence interval on the difference in mean life.

confidence level =99%

significance level =α= 0.01

Degree of freedom = n-1 = 8-1 =7

So, t_{\frac{\alpha}{2}}=3.499

Formula for confidence interval = \left( \bar{X} \pm t_{\frac{\alpha}{2}} \times \frac{s}{\sqrt{n}} \right)

Substitute the values

confidence interval = 939.5 \pm 3.499 \times \frac{1441.21}{\sqrt{8}} \right)

confidence interval = 939.5 - 3.499 \times \frac{1441.21}{\sqrt{8}} \right) to  = 939.5 + 3.499 \times \frac{1441.21}{\sqrt{8}} \right)

Confidence interval −843.396\ to  2722.396

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Answer:

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Step-by-step explanation:

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