What is the equation of this line? A: y=2x-3 B: y=1/3x-2 C: y=3x-2 D: y=1/2x-3

AnsweR quickkkkkKkkkkkk

Papessa [141]

Answer:

A & E

Step-by-step explanation:

_________________

4 0

2 years ago

the length of a rectangle is 5 centimeters less than six times its width. its area is 14 square centimeters. find the dimensions

V125BC [204]

Not sure if you are familiar with the quadratic equation, but see picture

3 0

3 years ago

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Convert 27 40 to a decimal using long division. A) 0.27 B) 0.675 C) 1.481 D) 1.5

ICE Princess25 [194]

Its B for sure not all of them just B lol

4 0

3 years ago

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Can someone explain this to me and how you do it. Also please help me get the answer and show some work on how you got it!

navik [9.2K]

Answer:

Step-by-step explanation:

$5\frac{1}{3}$ ÷ $\frac{4}{7}$ = $\frac{16}{3}$ ÷ $\frac{4}{7}$

= $\frac{16}{3}$ × $\frac{4}{7}$ = $\frac{64}{21}$

3 0

2 years ago

Use the Normal model N(1133,78) for the weights of steers.

Rudiy27

Answer:

Step-by-step explanation:

Let $X\sim N(1133, 78)$ .

Here, the mean is 1133 and standard deviation is 78.

Since we don't have the table for N(1133, 78), so we use the standard table for N(0,1),

a)

All we have to do is find, within the table the specific percentile.

now to find (42nd percentile) 0.42 in the normal table N(0,1) and extract the number on the row and column.

$P(X\leq x)=0.42$

Using: $Z=\frac{X-\mu}{\sigma}$ where $\mu$ is the mean and $\sigma$ is the standard deviation.

$P(\frac{X-1133}{78} \leq \frac{x-1133}{78})=0.42$

let $z=\frac{x-1133}{78}$

then, we have

$P(Z\leq z)=0.42$

Now, using Standard Normal table to find the value of z- score;

Usually, tables are set up as the probability that a number, z is less than or equal to Z.

i.e, $z= -0.20$

Now, putting this value in $z=\frac{x-1133}{78}$ , to find x;

$\frac{x-1133}{78}=-0.20$

On Simplify, we get;

$x=1,117.4$ pounds

b)

Similarly, find the weight for 91st percentile.

Follow the same steps that we have done in part (a),

$P(X\leq x)=0.91$ or

$P(\frac{X-1133}{78} \leq \frac{x-1133}{78})=0.91$

⇒ $P(Z \leq \frac{x-1133}{78})=0.91$

let $z=\frac{x-1133}{78}$

then, we have

$P(Z\leq z)=0.91$

Now, use normal table value to find the z-score;

Usually, tables are set up as the probability that a number, z is less than or equal to Z

we have, z= 1.34

putting the z value in $z=\frac{x-1133}{78}$ we get,

$\frac{x-1133}{78}=1.34$

On simplify, we get

x= 1,237.52 pounds

c)

the interquartile range (IQR)=3rd Quartile - 1st quartile

First find the 1st quartile:

$P(\frac{X-1133}{78} \leq \frac{x-1133}{78})=0.25$

⇒ $P(Z \leq \frac{x-1133}{78})=0.25$

let $z=\frac{x-1133}{78}$

then, we have

$P(Z\leq z)=0.25$

Now, we use normal table to find that z=-0.675

putting the z value in $z=\frac{x-1133}{78}$ we get,

$\frac{x-1133}{78}=-0.675$

On simplify, we get

x= 1,080.35 pounds

Similarly, for third quartile

By symmetry z=0.675

putting the z value in $z=\frac{x-1133}{78}$ we get,

$\frac{x-1133}{78}=0.675$

On simplify, we get

x= 1,185.65 pounds

then, IQR = 1185.65 -1080.35=105.3 pounds

7 0

3 years ago