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3 years ago

I REALLY NEED HELP ON 40 and 41, 50 POINTS

Mathematics

1 answer:

photoshop1234 [79]3 years ago

3 0

Question 40 is D because from looking at the graph, its is symmetrical and the mean is the average, which usually the middle of the graph, so is the median. The median and mean would be within the 2.1-3 category because its the middle of the graph.

Question 41 is B because every pound is 16 ounces. If you multiply 16 and 12.5, you'll get 200.

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If B is the midpoint of AC, and AB=5/2x+12 and AC=12x-4, what is the length of BC?

Setler [38]

Answer:

Step-by-step explanation:

We know that AC = 2 * AB so we can write:

12x - 4 = 2(5/2 x + 12)

12x - 4 = 5x + 24

7x = 28

x = 4

Since AB = BC the answer is 5/2 * 4 + 12 = 22.

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3 years ago

An employee earns a 12% commission rate on the cameras he sells. If he earns $384 in commission, what were his total sales?

Vlad1618 [11]

I do questions like these a little different than most people so...

384/12= 32 (32 is 1% of his sales)
32*100= 3200

His total sales were $3200.

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2 years ago

Read 2 more answers

Solve for x. a=1/2h(b x)

fredd [130]

A = 1/2 h(b + x)
2a = h(b + x)
2a/h = b + x
x = 2a/h - b

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3 years ago

The ratio of students to adults on a field trip is 8 to 1. Which table correctly shows this ratio for each grade?

iogann1982 [59]

The answer would be b hope this helps :)

5 0

2 years ago

P = √mx_ £x MX t make x the subject

vladimir2022 [97]

So, making x subject of the formula, x = [m - 2pt³ ±√(m² - 4pt²m)]/{2t⁵}

<h3>How to make x subject of the formula?</h3>

Since p = √(mx/t) - t²x

So, p + t²x = √(mx/t)

Squaring both sides, we have

(p + t²x)² = [√(mx/t)]²

p² + 2pt²x + t⁴x² = mx/t

Multiplying through by t,we have

(p² + 2pt²x + t⁴x²)t = mx/t × t

p²t + 2pt³x + t⁵x² = mx

p²t + 2pt³x + t⁵x² - mx = 0

t⁵x² + 2pt³x - mx + p²t = 0

t⁵x² + (2pt³ - m)x + p²t = 0

Using the quadratic formula, we find x.

$x = \frac{-b +/-\sqrt{b^{2} - 4ac} }{2a}$

where

a = t⁵,
b = (2pt³ - m) and
c = p²t

Substituting the values of the variables into the equation, we have

$x = \frac{-(2pt^{3} - m) +/-\sqrt{(2pt^{3} - m)^{2} - 4(t^{5})(p^{2}t) } }{2t^{5} }\\= \frac{-(2pt^{3} - m) +/-\sqrt{4p^{2} t^{6} - 4pt^{2}m + m^{2} - 4p^{2}t^{6} } }{2t^{5}}\\= \frac{-(2pt^{3} - m) +/-\sqrt{m^{2} - 4pt^{2}m } }{2t^{5}}\\= \frac{m - 2pt^{3} +/-\sqrt{m^{2} - 4pt^{2}m } }{2t^{5}}$

So, making x subject of the formula, $x = \frac{m - 2pt^{3} +/-\sqrt{m^{2} - 4pt^{2}m } }{2t^{5}}$

Learn more about subject of formula here:

brainly.com/question/25334090

#SPJ1

6 0

1 year ago