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3 years ago

12

In a probability experiment, Karen flipped a coin 84 times. The coin landed on heads 36 times. What percentage of the coin flips

resulted in tails? Round to the nearest percent.

2 answers:

hodyreva [135]3 years ago

8 0

Answer:

57.1%

Step-by-step explanation:

We know that a coin has only two sides, heads or tails. Therefore if of the 84 times, 36 times he fell in heads, the other times he fell in tails.

84 - 36 = 48 times fell in tails.

Therefore to calculate the percentage that fell in tails, it would be the ratio between the number of times it fell in tails and the total number of experiments:

48/84 = 0.571, and that is 57.1%

Which means that 57.1% of the times the coin was thrown fell into tails.

Vika [28.1K]3 years ago

3 0

Answer:

57%

Step-by-step explanation:

Total number of times the coin was flipped=84

Number of Times the Coin Landed on Heads=36

Therefore:

Total Number of Flips=Number of Heads + Number of Tails

84=36+Number of Tails

Number of Tails=84-36=48

We want to calculate what percentage of the coin flips landed on Tails

$=\frac{Number of tails}{Total Number of Trials} X100\\=\frac{48}{84}X100$

=57.14

The percentage of the coin flip that landed on Tails =57% ( to the nearest percent)

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2 years ago

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Hunter-Best [27]

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Step-by-step explanation:

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3 years ago

Write the equation of the quadratic function that passes through the points (-1, 1), (1, 5), and (2,10).

Serjik [45]

Answer:

$\displaystyle f(x)=x^2+2x+2$

Step-by-step explanation:

<u>System Of Linear Equations </u>

In this problem, we'll need to solve a 3x3 system of linear equations because we have three unknowns and three conditions.

We are required to find the equation of the quadratic function that passes through the points (-1, 1), (1, 5), and (2,10)

The general quadratic function can be written as

$\displaystyle f(x)=ax^2+bx+c$

We need to find the values of a,b, and c. Let's use the first condition, i.e. f(-1)=1

$\displaystyle f(-1)=a(-1)^2+b(-1)+c$

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Now we use the second condition f(1)=5

$\displaystyle f(1)=a(1)^2+b(1)+c$

$\displaystyle f(1)=a+b+c$

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Finally, we use the third condition f(2)=10

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$\displaystyle f(2)=4a+2b+c$

$\displaystyle 4a+2b+c=10....[eq\ 3]$

We put together eq 1, eq 2, and eq 3 to form the system

$\displaystyle \left\{\begin{matrix}a-b+c=1\\ a+b+c=5\\ 4a+2b+c=10\end{matrix}\right.$

Adding the first two equations we have

$\displaystyle 2a+2c=6$

$\displaystyle a+c=3$

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$\displaystyle b=2$

Using the above equation and the value of b in the third equation, we have

$\displaystyle \left\{\begin{matrix}a+c=3\\ 4a+c=6\end{matrix}\right.$

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6 0

3 years ago

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dem82 [27]

Answer:

$2x+ \frac{156}{5}$

Step-by-step explanation

simplify step-by-step.

2(x+7)−(−18+

4

5

)

Distribute the Negative Sign:

=2(x+7)+−1(−18+

4

5

)

=2(x+7)+(−1)(−18)+−1(

4

5

)

=2(x+7)+18+

−4

5

Distribute:

=(2)(x)+(2)(7)+18+

−4

5

=2x+14+18+

−4

5

Combine Like Terms:

=2x+14+18+

−4

5

=(2x)+(14+18+

−4

5

)

=2x+ $\frac{156}{5}$

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Answer:

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