All categories

2 years ago

Whose answer do you think is more likely to be representative of all the students in the schools? Explain.

Mathematics

1 answer:

maxonik [38]2 years ago

4 0

Answer:

Jordan's answer

Step-by-step explanation:

Because she went us to random people versus only the 8th grade kids

You might be interested in

What fraction of 2 kg is 275g? Give your answer in simplest form.

Katen [24]

Answer:

Step-by-step explanation:

1kg=1000g

2kg=2000g

275/2000=11/80g

4 0

3 years ago

Read 2 more answers

3. What is the last step to complete a perpendicular bisector construction? *

Furkat [3]

Its

C. Draw a line connecting the the intersection of the arcs below and above the segment.

6 0

3 years ago

It takes Claudia 1/4 of an hour to make a pizza. Claudia has 4 hours to make pizzas. How many pizzas can Claudia make during thi

juin [17]

Answer:

Step-by-step explanation:

divide 4 by .25 (or 1/4) and you get 16

7 0

2 years ago

Julio is paid 1.4 times his normal hourly rate for each hour he works over 32 hours in a week. Last week he worked 43 hours and

nekit [7.7K]

Answer:

$r = \$14.71$

Step-by-step explanation:

Given

Normal Hour = 32 hours

Overtime = Hours above 32

Rate for Overtime = 1.4 times normal rate

Earnings = $535.62

Required

Determine the normal hour pay

First, we need to determine the hours worked overtime.

This is:

$Overtime = 43 - 32$

$Overtime = 11$

The equation that binds all the parameters is:

$Earnings = Normal\ Hour * Normal\ Rate + Overtime * Overtime\ Rate$

This gives:

$535.62 = 21 * r + 11 * 1.4r$

$535.62 = 21r + 15.4r$

$535.62 = 36.4r$

Solve for r

$r = 535.62/36.4$

$r = \$14.71$

4 0

2 years ago

Consider the first five steps of the derivation of the Quadratic Formula.

lara31 [8.8K]

Answer:

Full proof below

Step-by-step explanation:

$\displaystyle ax^2+bx+c=0\\\\ax^2+bx=-c\\\\x^2+\biggr(\frac{b}{a}\biggr)x=-\frac{c}{a}\\\\x^2+\biggr(\frac{b}{a}\biggr)x+\biggr(\frac{b}{2a}\biggr)^2=-\frac{c}{a}+\biggr(\frac{b}{2a}\biggr)^2\\\\x^2+\biggr(\frac{b}{a}\biggr)x+\frac{b^2}{4a^2}=-\frac{c}{a}+\frac{b^2}{4a^2}\\ \\\biggr(x+\frac{b}{2a}\biggr)^2=-\frac{4ac}{4a^2}+\frac{b^2}{4a^2}\\ \\\biggr(x+\frac{b}{2a}\biggr)^2=\frac{b^2-4ac}{4a^2}\\ \\x+\frac{b}{2a}=\pm\frac{\sqrt{b^2-4ac}}{2a}\\ \\x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

6 0

1 year ago