Answer:
$35
Step-by-step explanation:
Rounding to the nearest dollar which would be 4 so in that case you would use the 5 in .57 and because 5 is a large number it would make the 4 into a five thus making it $35. Hope this helped :)
Answer:you need more information
Step-by-step explanation:you need more info like now!
Title:
<h2>See the explanation.</h2>
Step-by-step explanation:
The coin is to be flipped 2000 times.
The probability of getting a head is 0.3.
It is given that, we need to get a head in between 575 and 618 times.
If we will get 575 heads then we will get (2000 - 575) tails.
Hence, if we take that we want to get n times head, then we will get (2000 - n) times tail.
The probability of not getting a head is (1 - 0.3) = 0.7.
The probability of getting n times head is
.
The required probability is ∑
, where
.
The answer is: 3.91 inches .
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Note: Volume of cylinder: V = (base area) * (height);
in which: V = volume = 384 in.³ ;
h = height = 8 in. ;
Base area = area of the base (that is; "circle") = π r² ;
in which; "r" = radius;
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Solve for "r" :
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V = π r² * (8 in.) ;
384 in.³ = (8 in.) * (π r²) ;
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Divide EACH SIDE of the equation by "8" ;
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(384 in.³) / 8 = [ (8 in.) * (π r²) in.] / 8 ;
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to get:
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48 in.³ = (π r²) in.² * in. ;
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↔ (π r²) in.² * in. = 48 in.³ ;
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Rewrite this equation; using "3.14" as an approximation for: π ;
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(3.14 * r²) in.² * in. = 48 in.³
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Divide EACH SIDE of the equation by:
"[(3.14)*(in.²)*(in.)]" ; to isolate "r² " on one side of the equation;
(since we want to solve for "r") ;
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→ [(3.14 * r²) in.² * in.] / [(3.14)*(in.²)*(in.)] = 48 in.³ / [(3.14)*(in.²)*(in.)] ;
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→ to get: r² = 48/3.14 ;
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→ r² = 15.2866242038216561 ;
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To solve for "r" (the radius; take the "positive square root" of EACH side of the equation:
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→ +√(r²) = +√(15.2866242038216561)
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→ r = 3.9098112747064475286 ; round to 3.91 inches .
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Answer:
The answer for number 2 is 1/5.
Step-by-step explanation:
The fraction would be 4/20 but you can simplify it by dividing by 4.
4/4 is 1 and and 20/4 is 5.
you put 1 over 5 and that's how you get your answer.