You flip a coin and roll a regular six-sided number cube. Find the probability of flipping tails and rolling an even number

Mathematics

1 answer:

SpyIntel [72]3 years ago

6 0

Answer:

Since flipping a coin and rolling a 6-sided die are independent events ,thus the probability of getting head and an even number is just the product of the individual probability of events.

Let A be the event of getting a head ,and B be the event of getting an even number on die.

therefore, P(A and B)=P(A) P(B)=(1/2)*(3/6)=1/4=0.25

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6 0

2 years ago

Abdullah is kicking a soccer ball into the air. The path of the soccer ball can be modeled by a quadratic equation where the poi

kirill [66]

9514 1404 393

Answer:

A. 0 = −(x−6)^2+12

Step-by-step explanation:

The vertex form of a quadratic equation is ...

y = a(x -h)² +k . . . . . . for vertex (h, k) and scale factor 'a'

When the given vertex (h, k) = (6, 12) is used, we find the form of the equation to be ...

y = a(x -6)² +12

The ball will be on the ground when y = 0. Here, the vertical scale factor 'a' is -1, so the equation representing the ball being on the ground is ...

0 = -(x -6)² +12 . . . . . . matches choice A

5 0

3 years ago

Simplify the expression. (81x^12/3x^3)^1/3

ser-zykov [4K]

Raise each number to the power outside the parentheses.

$\left( \dfrac{81x^{12}}{3x^3} \right) ^{ \frac{1}{3} } = \\

= \left( 27x^{12-3} \right) ^{ \frac{1}{3} } = \\

= \left( 27x^{9} \right) ^{ \frac{1}{3} } = \\

= \left( 3^3x^{9} \right) ^{ \frac{1}{3} } = \\

= \left( 3^3 \right) ^{ \frac{1}{3} } \times \left( x^{9} \right) ^{ \frac{1}{3} } \\

= \left( 3^{3 \times \frac{1}{3} } \right) \times \left( x^{9 \times \frac{1}{3} }} \right) \\

= 3^1 x^3 \\

= 3x^3


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