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3 years ago

Jason rolls the die 14 times. What is the experimental probability that he will roll a 2

1 answer:

5 0

There’s 14 rolls. The probability he lands on a 2 would be 2/14. Once you simplify it it would give you a probability of 1/7

You might be interested in

Let f(x)=x2−2x−4 . What is the average rate of change for the quadratic function from x=−1 to x = 4?

goblinko [34]

Answer:

The average rate of change for f(x) from x=−1 to x = 4 is, 1

Step-by-step explanation:

Average rate A(x) of change for a function f(x) over [a, b] is given by:

$A(x) = \frac{f(b)-f(a)}{b-a}$

As per the statement:

$f(x) = x^2-2x-4$

we have to find the average rate of change from x = -1 to x = 4

At x = -1

$f(-1) = (-1)^2-2(-1)-4 = 1+2-4 = -1$

and

at x = 4

$f(4) = (4)^2-2(4)-4 = 16-8-4 = 4$

Substitute these in [1] we have;

$A(x) = \frac{f(4)-f(-1)}{4-(-1)}$

⇒ $A(x) = \frac{4-(-1)}{4+1}$

⇒ $A(x) = \frac{5}{5}$

Simplify:

A(x) = 1

Therefore, the average rate of change for f(x) from x=−1 to x = 4 is, 1

4 0

3 years ago

Read 2 more answers

Find the difference when the polynomial -5x^2+3x+8 is subtracted from the polynomial 2x^2+4x+1.

geniusboy [140]

The difference is $7x^{2} + 1x-7$ .

Step-by-step explanation:

Step 1:

The polynomial $-5x^{2}+3x+8$ is subtracted from the polynomial $2x^{2} + 4x +1$ .

If we write this as an equation, we get

$(2x^{2} + 4x +1) -(-5x^{2}+3x+8)$ .

To subtract the polynomials, we group up the terms based on their variables.

In this subtraction, there are two variables i.e. $x^{2}$ and $x$ and there is one constant term.

Step 2:

The subtraction of the $x^{2}$ variables; $2x^{2} -(-5x^{2} ) = 2x^{2} +5x^{2} = 7x^{2}$ .

The subtraction of the $x$ variables; $4x - (3x) = 1x$ .

The subtraction of the constants; $1-(8) = -7$ .

So $(2x^{2} + 4x +1) -(-5x^{2}+3x+8) = 7x^{2} + 1x-7$ .

4 0

3 years ago

-3+m/3=12 how do I solve this

timama [110]

First: You add 3 to each side. Second: You multiply each side by 3. Third: You gaze in wondrous amazement at the answer, which lies revealed before you on the page.

3 0

3 years ago

Read 2 more answers

Remember to show work and explain. Use the math font.

MrMuchimi

Answer:

$\large\boxed{1.\ f^{-1}(x)=4\log(x\sqrt[4]2)}\\\\\boxed{2.\ f^{-1}(x)=\log(x^5+5)}\\\\\boxed{3.\ f^{-1}(x)=\sqrt{4^{x-1}}}$

Step-by-step explanation:

$\log_ab=c\iff a^c=b\\\\n\log_ab=\log_ab^n\\\\a^{\log_ab}=b\\\\\log_aa^n=n\\\\\log_{10}a=\log a\\=============================$

$1.\\y=\left(\dfrac{5^x}{2}\right)^\frac{1}{4}\\\\\text{Exchange x and y. Solve for y:}\\\\\left(\dfrac{5^y}{2}\right)^\frac{1}{4}=x\qquad\text{use}\ \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}\\\\\dfrac{(5^y)^\frac{1}{4}}{2^\frac{1}{4}}=x\qquad\text{multiply both sides by }\ 2^\frac{1}{4}\\\\\left(5^y\right)^\frac{1}{4}=2^\frac{1}{4}x\qquad\text{use}\ (a^n)^m=a^{nm}\\\\5^{\frac{1}{4}y}=2^\frac{1}{4}x\qquad\log_5\ \text{of both sides}$

$\log_55^{\frac{1}{4}y}=\log_5\left(2^\frac{1}{4}x\right)\qquad\text{use}\ a^\frac{1}{n}=\sqrt[n]{a}\\\\\dfrac{1}{4}y=\log(x\sqrt[4]2)\qquad\text{multiply both sides by 4}\\\\y=4\log(x\sqrt[4]2)$

$--------------------------\\2.\\y=(10^x-5)^\frac{1}{5}\\\\\text{Exchange x and y. Solve for y:}\\\\(10^y-5)^\frac{1}{5}=x\qquad\text{5 power of both sides}\\\\\bigg[(10^y-5)^\frac{1}{5}\bigg]^5=x^5\qquad\text{use}\ (a^n)^m=a^{nm}\\\\(10^y-5)^{\frac{1}{5}\cdot5}=x^5\\\\10^y-5=x^5\qquad\text{add 5 to both sides}\\\\10^y=x^5+5\qquad\log\ \text{of both sides}\\\\\log10^y=\log(x^5+5)\Rightarrow y=\log(x^5+5)$

$--------------------------\\3.\\y=\log_4(4x^2)\\\\\text{Exchange x and y. Solve for y:}\\\\\log_4(4y^2)=x\Rightarrow4^{\log_4(4y^2)}=4^x\\\\4y^2=4^x\qquad\text{divide both sides by 4}\\\\y^2=\dfrac{4^x}{4}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\y^2=4^{x-1}\Rightarrow y=\sqrt{4^{x-1}}$