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3 years ago

Can someone please help me with number 19

Mathematics

1 answer:

kipiarov [429]3 years ago

3 0

The expression of the problem is, 8x.
where "x" stands for the number of dogs. so you would then substitute x with 12. Then you would solve the problem to get your answer.(8•12)

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*WILL GIVEE BRAINLIEST FOR THE BEST ANSWER* Which of the following is NOT a biased sample?

Dafna1 [17]

Answer:

Step-by-step explanation:

All other options are more biased

In B, the official wants to use 50 people from a library to represent the entire town's opinion on the library itself.... that's biased.

In C, it's the same. Of course basketball players would like to watch basketball.... right?? That's why the choice is biased.

In D, the location is a deli.... and from experience people order meat at a deli... cuz why else go there?

8 0

2 years ago

39. You need to rent a car for a trip. Wrecko'mend charges $. 10 a mile plus a $30 flat fee. Rent a Wreck

Wewaii [24]

i think this is a systems of equations

putting this on a graph we have to equations:

y = 10x + 30

y = 20x

graph and find the intersection.

it has to be more than or equal to 3 miles i think for wreckomend

5 0

2 years ago

You own 3/5 of the coins in a collection. What percent of the coins do you own?

Goryan [66]

60%
You divide 3/5 and then you get a decimal 0.6. Then you convert it to a percentage

4 0

2 years ago

Read 2 more answers

What is the third quartile of the data set? 23, 35, 55, 61, 64, 67, 68, 71, 75, 94, 99

lara [203]

Answer:

Step-by-step explanation:

Q1=55 Q2=67 Q3=75

4 0

2 years ago

a line whose perpendicular distance from the origin is 4 units and the slope of perpendicular is 2÷3. Find the equation of the l

GrogVix [38]

Answer:

$\huge\boxed{y=\dfrac{2}{3}x-\dfrac{4\sqrt{13}}{3}\ \vee\ y=\dfrac{2}{3}x+\dfrac{4\sqrt{13}}{3}}$

Step-by-step explanation:

The equation of a line:

$y=mx+b$

We have

$m=\dfrac{2}{3}$

substitute:

$y=\dfrac{2}{3}x+b$

The formula of a distance between a point and a line:

General form of a line:

$Ax+By+C=0$

Point:

$(x_0,\ y_0)$

Distance:

$d=\dfrac{|Ax_0+By_0+C|}{\sqrt{A^2+b^2}}$

Convert the equation:

$y=\dfrac{2}{3}x+b$ |subtract $y$ from both sides

$\dfrac{2}{3}x-y+b=0$ |multiply both sides by 3

$2x-3y+3b=0\to A=2,\ B=-3,\ C=3b$

Coordinates of the point:

$(0,\ 0)\to x_0=0,\ y_0=0$

substitute:

$d=4$

$4=\dfrac{|2\cdot0+(-3)\cdot0+3b|}{\sqrt{2^2+(-3)^2}}\\\\4=\dfrac{|3b|}{\sqrt{4+9}}$

$4=\dfrac{|3b|}{\sqrt{13}}\qquad|$ |multiply both sides by $\sqrt{13}$

$4\sqrt{13}=|3b|\iff3b=-4\sqrt{13}\ \vee\ 3b=4\sqrt{13}$ |divide both sides by 3

$b=-\dfrac{4\sqrt{13}}{3}\ \vee\ b=\dfrac{4\sqrt{13}}{3}$

Finally:

$y=\dfrac{2}{3}x-\dfrac{4\sqrt{13}}{3}\ \vee\ y=\dfrac{4\sqrt{13}}{3}$

4 0

2 years ago