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4 years ago

Simplify

Mathematics

1 answer:

faust18 [17]4 years ago

4 0

Answer:What are the following?

Step-by-step explanation:

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Write the equation in slope-intercept form of the line that has a slope of 6 and contains the point (1, 1).

valentinak56 [21]

Answer: y=6x-5

Step-by-step explanation:

Since we know the slope, we can plug it into the slope-intercept equation. We can plug in the given point to find the y-intercept.

y=6x+b

1=6(1)+b

1=6+b

b=-5

Now that we know the y-intercept, we can complete the equation.

y=6x-5

5 0

3 years ago

What is the average (arithmetic mean) of the first 90 measurements in a certain list of 92 measurements? (1) the average of the

marishachu [46]

The sum of all the measurements is just the average times the number.

$\bar x = \frac{1}{n} \sum_{i=1}^n x_i$

$\sum_{i=1}^n x_i = n \bar x$

So if the average of all 92 is 7 the sum of those is $92 \times 7$

The average of the last two is 7.2 so their sum is $2 \times 7.2$

That means the sum of the first 90 is $92 \times 7 - 2 \times 7.2$

so the average of the first 90 is

$\dfrac{92 \times 7 - 2 \times 7.2}{90} = 6.99556$ cm

7 0

3 years ago

....Help me please :)))))))

My name is Ann [436]

I’m pretty sure it’s just 1/8

7 0

3 years ago

P(k)=a^k=2 3 4 find value of a that makes this is a valid probability distribution

Vesna [10]

Sounds like you're asked to find $a$ such that

$\displaystyle\sum_{k=2}^4\mathbb P(k)=\mathbb P(2)+\mathbb P(3)+\mathbb P(4)=1$

In other words, find $a$ that satisfies

$a^2+a^3+a^4=1$

We can factorize this as

$a^4+a^3+a^2-1=a^3(a+1)+(a-1)(a+1)=(a+1)(a^3+a-1)=0$

In order that $\mathbb P(k)$ describes a probability distribution, require that $\mathbb P(k)\ge0$ for all $k$ , which means we can ignore the possibility of $a=-1$ .

Let $a=y+\dfrac xy$ .

$a^3+a-1=\left(y+\dfrac xy\right)^3+\left(y+\dfrac xy\right)-1=0$
$\left(y^3+3xy+\dfrac{3x^2}y+\dfrac{x^3}{y^3}\right)+\left(y+\dfrac xy\right)-1=0$

Multiply both sides by $y^3$ .

$y^6+3xy^4+3x^2y^2+x^3+y^4+xy^2-y^3=0$

We want to find $x\neq0$ that removes the quartic and quadratic terms from the equation, i.e.

$\begin{cases}3x+1=0\\3x^2+x=0\end{cases}\implies x=-\dfrac13$

so the cubic above transforms to

$y^6-y^3-\dfrac1{27}=0$

Substitute $y^3=z$ and we get

$z^2-z-\dfrac1{27}=0\implies z=\dfrac{9+\sqrt{93}}{18}$
$\implies y=\sqrt[3]{\dfrac{9+\sqrt{93}}{18}}$
$\implies a=\sqrt[3]{\dfrac{9+\sqrt{93}}{18}}-\dfrac13\sqrt[3]{\dfrac{18}{9+\sqrt{93}}}$

6 0

4 years ago

13.12 / .10 equals what please help me

sleet_krkn [62]

Answer:

131.2

Step-by-step explanation:

6 0

3 years ago

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