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melisa1 [442]
2 years ago
8

Please help!!!

Mathematics
1 answer:
Gala2k [10]2 years ago
6 0
Vsphere=(4/3)pi(r)³

remember that d/2=r, so 2r=d

given
V=2400

2400=(4/3)pi(r)³
times both sides by (3/4)
1800=pi(r)³
divide both sides by pi
1800/pi=r³
cube root both sides
<span>8.30566118415=r
times 2
</span><span>16.6113223683=r</span>
the aproximate diameter isi 16.6 cm
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What is the sum of 5.89732 304.615
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3 years ago
Can someone please answer 5, 6 and 7? Thank you.
Montano1993 [528]
Reduce a 24 cm by 36 cm photo to 3/4 original size.

The most logical way to do this is to keep the width-to-height ratio the same:  It is 24/36, or 2/3.  The original photo has an area of (24 cm)(36 cm) = 864 cm^2.

Let's reduce that to 3/4 size:  Mult. 864 cm^2 by (3/4).  Result:  648 cm^2.

We need to find new L and new W such that W/L = 2/3 and WL = 648 cm^2.

From the first equation we get W = 2L/3.  Thus, WL = 648 cm^2 = (2L/3)(L).

Solve this last equation for L^2, and then for L:

2L^2/3 = 648, or (2/3)L^2 = 648.  Thus, L^2 = (3/2)(648 cm^2) = 972 cm^2.

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6 0
3 years ago
In a metal fabrication​ process, metal rods are produced to a specified target length of 15 feet. Suppose that the lengths are n
Leto [7]

Answer:

95% Confidence interval: (14.4537 ,15.1463)

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 15 feet

Sample mean, \bar{x} = 14.8 feet

Sample size, n = 16

Alpha, α = 0.05

Sample standard deviation, σ = 0.65 feet

Degree of freedom = n - 1 = 15

95% Confidence interval:

\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}  

Putting the values, we get,  

t_{critical}\text{ at degree of freedom 15 and}~\alpha_{0.05} = \pm 2.1314  

14.8 \pm 2.1314(\dfrac{0.65}{\sqrt{16}} ) \\\\= 14.8 \pm 0.3463 = (14.4537 ,15.1463)  

is the required confidence interval for the true mean length of rods.

3 0
3 years ago
Read 2 more answers
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