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3 years ago

PLS HELP ME WITH THISS THANK YOU !

Mathematics

2 answers:

Fantom [35]3 years ago

4 0

Answer:

The answer is C)5q⁶r⁸

Step-by-step explanation:

Hope this helps

Tanya [424]3 years ago

3 0

Yep what the other person said .

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How do you simplify this expression? -12 divided by 3•(-8+(-4)^2-6)+2

stepan [7]

I good accumulator to use would be symbolab.com

The answer would be 3072/5629

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3 years ago

I don't remember how to find x please help

alexandr1967 [171]

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2 years ago

Katie is making covers for metal nameplates. She needs to know how much plastic, at minimum, she’ll need to fully cover 30 namep

grigory [225]

1404 is the surface area of 30 nameplates.

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3 years ago

Suppose that 3 balls will be randomly put into 3 buckets, with each ball being equally likely to be put into each of the buckets

Yuliya22 [10]

Answer:

$\frac{1}{27}$

Step-by-step explanation:

Since there are a total of three buckets and only one can be chosen at a time, this would mean that the probability of a ball being placed in a bucket is 1/3. Since each ball has the same probability of being placed into any bucket regardless of the where the previous ball landed, it means that each ball has the same 1/3 probability of a bucket. In order to find the probability that all three land in the same bucket, we need to multiply this probability together for each one of the balls like so...

$\frac{1}{3} * \frac{1}{3} * \frac{1}{3} = \frac{1}{27}$

Finally, we see that the probability of all three balls landing in the same bucket is $\frac{1}{27}$

8 0

2 years ago

Para reunir dinero para su gira de estudios , los alumnos de un curso deciden vender números de una rifa que se encuentran numer

QveST [7]

Respuesta:

0.53

Explicación:

Para calcular la posibilidad del evento A: "ganar la rifa comprando todos los números múltiplos de 3 o 5", debemos usar la siguiente fórmula.

P(A) = casos favorables / casos posibles

Evaluemos primero todos los casos que son múltiplos de 3, entre 1 y 100: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99. En total son 33.

Ahora, evaluemos todos los casos que son múltiplos de 3, entre 1 y 100: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100. En total son 20.

El número total de casos favorables es 33 + 20 = 53.

El número de casos posibles es el total de números de 1 a 100, es decir 100.

Luego P(A) = 53/100 = 0.53.

7 0

2 years ago