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2 years ago

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What is the probability of tossing at least 1 tail of you toss 3 coins at once

1 answer:

Sedaia [141]2 years ago

5 0

Explanation: Probability of NOT getting a tail in 3 coin toss is (12)3=18 . Probability of getting at least 1 tail in 3 coin toss is 1−18=78

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Suppose the system of linear equations has no solution. Which is NOT a possible value of the constant for the second equation?

Tom [10]

Answer:

A) 5

Step-by-step explanation:

The coefficients of the second equation are 1/3 of those of the first equation. If the constant of the second equation is also 1/3 of the constant of the first equation, then the equations are <em>dependent</em>, and will have an infinite number of solutions.

The constant must be something other than 1/3·15 = 5 to make the equations <em>inconsistent</em>, having no solutions.

4 0

2 years ago

The function f(x) = x2 + 10x – 3 written in vertex form is f(x) = (x + 5)2 – 28. What are the coordinates of the vertex?

MrRa [10]

The vertex of the quadratic equation is (-5, - 28). In vertex form, y = a(x - h)2<span> + </span><span>k, (h, k) is the vertex of the equation.</span>

5 0

2 years ago

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A cube has a side length of 4 feet. What is the volume of the cube?

riadik2000 [5.3K]

Answer:

64 square ft^3

Step-by-step explanation:

If the side lengh is 3 then you would have 4^3 which is 64.

5 0

2 years ago

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Erin calculated the average speed of her most recent run to be 120/16 miles per hour how is this friction expressed as a decimal

Contact [7]

Answer:

$\dfrac{15}{2}$

Step-by-step explanation:

The average speed of the Erin is 120/16 mph

Here,

Numerator = 120

Denominator = 16

mph = miles per hour

We need to find an expression expressed as a decimal.

So,

$\dfrac{120}{16}=\dfrac{2\times 3\times 2\times 2\times 5}{2\times 2\times 2\times 2}\\\\=\dfrac{15}{2}$

So, the final answer is $\dfrac{15}{2}$ .

4 0

2 years ago

Given that 'n' is any natural numbers greater than or equal 2. Prove the following Inequality with Mathematical Induction

Oliga [24]

The base case is the claim that

$\dfrac11 + \dfrac12 > \dfrac{2\cdot2}{2+1}$

which reduces to

$\dfrac32 > \dfrac43 \implies \dfrac46 > \dfrac86$

which is true.

Assume that the inequality holds for <em>n</em> = <em>k </em>; that

$\dfrac11 + \dfrac12 + \dfrac13 + \cdots + \dfrac1k > \dfrac{2k}{k+1}$

We want to show if this is true, then the equality also holds for <em>n</em> = <em>k</em> + 1 ; that

$\dfrac11 + \dfrac12 + \dfrac13 + \cdots + \dfrac1k + \dfrac1{k+1} > \dfrac{2(k+1)}{k+2}$

By the induction hypothesis,

$\dfrac11 + \dfrac12 + \dfrac13 + \cdots + \dfrac1k + \dfrac1{k+1} > \dfrac{2k}{k+1} + \dfrac1{k+1} = \dfrac{2k+1}{k+1}$

Now compare this to the upper bound we seek:

$\dfrac{2k+1}{k+1} > \dfrac{2k+2}{k+2}$

because

$(2k+1)(k+2) > (2k+2)(k+1)$

in turn because

$2k^2 + 5k + 2 > 2k^2 + 4k + 2 \iff k > 0$

6 0

2 years ago

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