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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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8 0
3 years ago
Jackson is playing a game of chance where he must randomly draw a marble out of a jar. He calculates the probability of drawing
Sindrei [870]

Using complementary events, considering a \frac{1}{4} probability of drawing a red marble, the probability of not drawing a red marble is of \frac{3}{4}.

<h3>What are complementary events?</h3>

They are mutually exclusive events which have the sum of their probabilities as 1.

In this problem, we consider that there is a \frac{1}{4} probability of drawing a red marble, and since drawing a red marble and drawing a non-red marble are complementary events, we have that:

\frac{1}{4} + p = 1

p = \frac{3}{4}

The probability of not drawing a red marble is of \frac{3}{4}.

More can be learned about complementary events at brainly.com/question/9752956

8 0
2 years ago
The course grade in a statistics class is the average of the scores on five examinations. Suppose that a student's scores on the
nadya68 [22]

Answer:

84 is the highest possible course average

Step-by-step explanation:

Total number of examinations = 5

Average = sum of scores in each examination/total number of examinations

Let the score for the last examination be x.

Average = (66+78+94+83+x)/5 = y

5y = 321+x

x = 5y -321

If y = 6, x = 5×6 -321 =-291.the student cannot score -291

If y = 80, x = 5×80 -321 =79.he can still score higher

If If y = 84, x = 5×84 -321 =99.This would be the highest possible course average after the last examination.

If y= 100

The average cannot be 100 as student cannot score 179(maximum score is 100)

8 0
3 years ago
Can you answer the question in the picture below?
uysha [10]

Step-by-step explanation:

Area of parking lot

= length \: of \: base \:  \times height \\  = 200 \times 120 \\  = 24000 \:  {ft}^{2}

5 0
3 years ago
F(1) = 4, f(n) = f(n<br> − 1) + 11
zvonat [6]

Answer:

Step-by-step explanation:

You don't specify what you're supposed to do, so I'll make an educated guess.  

Given the sequence f(1) = 4, f(n) = f(n  − 1) + 11, find the first 5 terms:

f(1) = 4

f(2) = f(2 - 1) + 11 = f(1) + 11 = 4 + 11 = 15

f(3) = f(2) + 11 = 15 + 11 = 26

f(4) = f(3) + 11 = 26 + 11 = 37

f(5) = 37 + 11 = 48

4 0
3 years ago
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