A triangle has four sides and a dog has four legs. always sometimes never

What is the domain of the relation {(2, 8), (0, 8), (–1, 5), (–1, 3), (–2, 3)}?

lesya [120]

(2,0,-1,-1,-2) that is the domain from what I remember. Hope that help

6 0

3 years ago

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This unit focused, in part, on relationships in triangles. A triangle is the simplest type of

Elena-2011 [213]

Answer:

triangles have a specific geometric shape.Because they have three sides, the pythagorean theorem for right angled triangle.

7 0

2 years ago

There is a sales tax of $21 on an item that cost $217 before tax. A second item.cost $93 before tax.What is the sales tax on the

Lana71 [14]

Answer:

9

Step-by-step explanation:

its difficult to explain but I found that there is a 10.3%Tax so 10.3% of 93 is 9

7 0

3 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

4.08 dived by 2 show work

Lyrx [107]

Answer:

$4.08 \div 2 \\ 2.08 \: answer$

please mark me brainliest please

3 0

2 years ago

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