What Is the value of a ? Help Me ：）

How do you graph a line written in standard form

Sonbull [250]

Ok so standard form is a bit tricky, but I’ll help explain

Here’s and example:

If you have 2x+5y=10 here is how you would solve it

First you divide the x value by what the number is equal to (in our example, 10), and then we get 10/2 is 5. So now we know that

x is equal to five

Keep that in mind. Now we just have to do the same thing for the y value. 5/10 is 2, so you know that

y is equal to two

NOW you put those two together, and graph on the intercepts (the positions on the y and x axis where the lines intercept)

You would put a line on the x axis at 5, and on the y axis at 2

If you have any questions, you can ask

4 0

3 years ago

If we reject the null hypothesis, h0: ρ = 0 , what can we conclude about the population correlation coefficient?

ehidna [41]

If the null hypothesis, $H_0:\:\rho =0$ is rejected, then we can conclude that the correlation coefficient is significant. This means that there is enough evidence to conclude that a relationship exists between the two (or more) variables involved in the regression.

5 0

3 years ago

Below is the graph of y=b^x. Find b.

omeli [17]

Y = <span>b^x

when x = 1
y = b^1
y = b

Therefore, the value of b is the same as the value of y when x =1
From the graph,
When x = 1, y = 0.5
Therefore, b = 0.5

To confirm this

From the graph,

When x = -1, y = 2
Since </span>y = b^x<span>
2 = </span>b^-1
2 = 1/b
2b = 1
b = 0.5

When x = -2, y = 4
Since y = b^x
4 = b^-2
4 = 1/(b^2)
b^2 = 1/4
b = √(1/4)
b = 1/2
b = 0.5

Therefore, it is conformed that b = 0.5

5 0

3 years ago

Read 2 more answers

An area is approximated to be 14 in 2 using a left-endpoint rectangle approximation method. A right- endpoint approximation of t

USPshnik [31]

The trapezoidal approximation will be the average of the left- and right-endpoint approximations.

Let's consider a simple example of estimating the value of a general definite integral,

$\displaystyle\int_a^bf(x)\,\mathrm dx$

Split up the interval $[a,b]$ into $n$ equal subintervals,

$[x_0,x_1]\cup[x_1,x_2]\cup\cdots\cup[x_{n-2},x_{n-1}]\cup[x_{n-1},x_n]$

where $a=x_0$ and $b=x_n$ . Each subinterval has measure (width) $\dfrac{a-b}n$ .

Now denote the left- and right-endpoint approximations by $L$ and $R$ , respectively. The left-endpoint approximation consists of rectangles whose heights are determined by the left-endpoints of each subinterval. These are $\{x_0,x_1,\cdots,x_{n-1}\}$ . Meanwhile, the right-endpoint approximation involves rectangles with heights determined by the right endpoints, $\{x_1,x_2,\cdots,x_n\}$ .

So, you have

$L=\dfrac{b-a}n\left(f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1})\right)$
$R=\dfrac{b-a}n\left(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n)\right)$

Now let $T$ denote the trapezoidal approximation. The area of each trapezoidal subdivision is given by the product of each subinterval's width and the average of the heights given by the endpoints of each subinterval. That is,

$T=\dfrac{b-a}n\left(\dfrac{f(x_0)+f(x_1)}2+\dfrac{f(x_1)+f(x_2)}2+\cdots+\dfrac{f(x_{n-2})+f(x_{n-1})}2+\dfrac{f(x_{n-1})+f(x_n)}2\right)$

Factoring out $\dfrac12$ and regrouping the terms, you have

$T=\dfrac{b-a}{2n}\left((f(x_0)+f(x_1)+\cdots+f(x_{n-2})+f(x_{n-1}))+(f(x_1)+f(x_2)+\cdots+f(x_{n-1})+f(x_n))\right)$

which is equivalent to

$T=\dfrac12\left(L+R)$

and is the average of $L$ and $R$ .

So the trapezoidal approximation for your problem should be $\dfrac{14+21}2=\dfrac{35}2=17.5\text{ in}^2$

4 0

3 years ago

the temperature of an oven went through 330 degrees to 390 degrees what is the percentage increase in temperature

valentina_108 [34]

All you do is divide.
330/390= 0.846
round it: 0.85
Answer is about 85%

5 0

3 years ago