Responder:
La pizza con un perímetro de 100 cm es más grande ¿verdad?
Explicación paso a paso:
Deja que la pizza tenga forma circular.
Sea el área de la pizza = πd² / 4 y;
Perímetro de la pizza = πd
d es el diámetro de la pizza
Si la madre dice que el que tiene un perímetro de 100 cm es más grande, para estar seguros necesitamos obtener el diámetro de la pizza. La de mayor diámetro será la pizza más grande.
P = 100cm
100 = πd
d = 100 / π
d = 100 / 3,14
d = 31,85 cm
El diámetro de la pizza mamá es de 31,85 cm.
Si el padre dice que el que tiene un área de 100 cm² es más grande, obtengamos también el diámetro para estar seguros.
A = πd² / 4
100 = πd² / 4
400 = πd²
d² = 400 / π
d² = 400 / 3,14
d² = 127,39
d = √127,39
d = 11,29 cm
Por lo tanto, el diámetro de la pizza padre es de 11,29 cm.
Dado que el diámetro de la pizza madre es mayor que el de esa, la pizza con un perímetro de 100 cm es más grande, lo que demuestra que la madre tiene razón.
<h3>
Answers:</h3>
The first ordered pair is ( -4 , -3 )
The second ordered pair is ( 8, 3 )
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Explanation:
The first point is (x,-3) where x is unknown. It pairs up with y = -3 so we can use algebra to find x
x-2y = 2
x-2(-3) = 2 ... replace every y with -3; isolate x
x+6 = 2
x = 2-6
x = -4
The first point is (-4, -3)
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We'll do something similar for the other point. This time we know x but don't know y. Plug x = 8 into the equation and solve for y
x-2y = 2
8-2y = 2
-2y = 2-8
-2y = -6
y = -6/(-2)
y = 3
The second point is (8, 3)
<span>False. Perimeter is a linear measure. Square units are for area.</span>
Step-by-step explanation:
<h2>
<em><u>You can solve this using the binomial probability formula.</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows:</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: </u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) </u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157when x=3 (4 3)(1/6)^3(5/6)^4-3 = 0.0154</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157when x=3 (4 3)(1/6)^3(5/6)^4-3 = 0.0154when x=4 (4 4)(1/6)^4(5/6)^4-4 = 0.0008</u></em></h2><h2>
<em><u>You can solve this using the binomial probability formula.The fact that "obtaining at least two 6s" requires you to include cases where you would get three and four 6s as well.Then, we can set the equation as follows: P(X≥x) = ∑(k=x to n) C(n k) p^k q^(n-k) n=4, x=2, k=2when x=2 (4 2)(1/6)^2(5/6)^4-2 = 0.1157when x=3 (4 3)(1/6)^3(5/6)^4-3 = 0.0154when x=4 (4 4)(1/6)^4(5/6)^4-4 = 0.0008Add them up, and you should get 0.1319 or 13.2% (rounded to the nearest tenth)</u></em></h2>
