I am confused on this question.....please help!

What is the inverse operations of -5x + 10 = 40

noname [10]

Answer:

x=-6

Step-by-step explanation:

−5x+10=40

−5x+10−10=40−10

−5x=30

−5x/5=30/5

x=−6

6 0

3 years ago

The prices of all college textbooks follow a bell-shaped distribution with a mean of $113 and a standard deviation of $12. Using

Soloha48 [4]

Step-by-step explanation:

In statistics, the empirical rule states that for a normally distributed random variable,

68.27% of the data lies within one standard deviation of the mean.

95.45% of the data lies within two standard deviations of the mean.

99.73% of the data lies within three standard deviations of the mean.

In mathematical notation, as shown in the figure below (for a standard normal distribution), the empirical rule is described as

$\Phi(\mu \ - \ \sigma \ \leq X \ \leq \mu \ + \ \sigma) \ = \ 0.6827 \qquad (4 \ \text{s.f.}) \\ \\ \\ \Phi(\mu \ - \ 2\sigma \ \leq X \ \leq \mu \ + \ 2\sigma) \ = \ 0.9545 \qquad (4 \ \text{s.f.}) \\ \\ \\ \Phi}(\mu \ - \ 3\sigma \ \leq X \ \leq \mu \ + \ 3\sigma) \ = \ 0.9973 \qquad (4 \ \text{s.f.})$

where the symbol $\Phi$ (the uppercase greek alphabet phi) is the cumulative density function of the normal distribution, $\mu$ is the mean and $\sigma$ is the standard deviation of the normal distribution defined as $N(\mu, \ \sigma)$ .

According to the empirical rule stated above, the interval that contains the prices of 99.7% of college textbooks for a normal distribution $N(113, \ 12)$ ,

$\Phi(113 \ - \ 3 \ \times \ 12 \ \leq \ X \ \leq \ 113 \ + \ 3 \ \times \ 12) \ = \ 0.9973 \\ \\ \\ \-\hspace{1.75cm} \Phi(113 \ - \ 36 \ \leq \ X \ \leq \ 113 \ + \ 36) \ = \ 0.9973 \\ \\ \\ \-\hspace{3.95cm} \Phi(77 \ \leq \ X \ \leq \ 149) \ = \ 0.9973$

Therefore, the price of 99.7% of college textbooks falls inclusively between $77 and $149.

5 0

2 years ago

Can anyone help me to understand how to solve coordinate proofs in Geometry?????

nataly862011 [7]

The coordinate proof is a proof of a geometric theorem which uses "generalized" points on the Cartesian Plane to make an argument.The method usually involves assigning variables to the coordinates of one or more points, and then using these variables in the midpoint or distance formulas .For example, the following is a coordinate proof of the Triangle Midsegment Theorem , which states that the segment connecting the midpoints of two sides of a triangle is parallel to the third side and exactly half the length.Without loss of generality, we can assume that one side of the triangle lies on the xx -axis with one vertex at (0,0)(0,0) and the other vertex at(a,0)(a,0).Let the third vertex have the coordinates (b,c)(b,c) . We can assume without loss of generality that this third vertex lies in the first quadrant. (If it doesn't, you can just reflect the triangle over the xx - and/or yy -axis until it does.)

Example:

To Prove: XY¯¯¯¯¯¯XY¯ is parallel to PR¯¯¯¯¯PR¯ and

XY=12PRXY=12PR

Proof:

Using the midpoint formula , the coordinates of the midpoints XX and YY are

X(0+b2,0+c2)=X(b2,c2)Y(a+b2,0+c2)=Y(a+b2,c2)X(0+b2,0+c2)=X(b2,c2)Y(a+b2,0+c2)=Y(a+b2,c2)

First note that the slope is 00 , since

m=riserun =(c2−c2)(b2−(a+b2)) =0(b2−(a+b2)) =0m=riserun =(c2−c2)(b2−(a+b2)) =0(b2−(a+b2)) =0

So, XY¯¯¯¯¯¯XY¯ is parallel to PR¯¯¯¯¯PR¯ .

Now, compare the lengths of the two segments.

XY=a+b2−b2 =a2PR=a−0 =aXY=a+b2−b2 =a2PR=a−0 =a

So,