Mariana writes the equation below to represent a problem. Which problem could the

Mathematics

1 answer:

Nataly_w [17]2 years ago

6 0

Answer:

Mariana uses 6 boxes of fertilizer each box weighs 15 ounces How much fertilizer does she use in all

Step-by-step explanation:

You might be interested in

Using the standard normal table, the probability that Seth’s light bulb will last no more than 14,500 (P(z ≤ 1.25)) hours is abo

lana66690 [7]

Answer:

How many standard deviations above the mean is 14,500 hours? 1.25 1.5 2.5 Using the standard normal table, the probability that Seth's light bulb will last no more than 14,500 (P(z ≤ 1.25)) hours is about ✔ 89% .

5 0

1 year ago

Read 2 more answers

Byron has 1 7/10 kilograms of black pepper. He uses 7/8 of the black pepper and splits between 7 pepper shakers. How much pepper

IrinaVladis [17]

Amount of black pepper = 1 7/10 kg = 17/10 kg.
Amount used is
(7/8)*(17/10) = 119/80 kg

This amount is split between 7 shakers. Each shaker has
(119/80)/7 = 17/80 kg = 0.2125 kg

Answer: 17/80 kg (or 0.2125 kg)

4 0

1 year ago

Find the area of the shaded portion in the equilateral triangle with sides 6. Show all work for full credit. (Hint: Assume that

Vadim26 [7]

So... if you notice the picture below

each circle, has their central angle at the vertex of the triangle
that simply means, 3 circles with a radius of 3, overlapping the triangle

now, the area in the middle, the shaded one, will be, the whole area of the triangle MINUS those 3 circle sectors

hmmmm each sector has 60°, that means, all three of them will then be 60+60+60 or 180°, so the area of those three sectors, can be combined into a 180° sector, well, hell, 180° is really half a circle

so.... the area of those three sectors of 60° each, all three combined, is the same area of half a circle with a radius of 3

so $\bf \textit{area of an equilateral triangle}\\\\
A=\cfrac{s^2\sqrt{3}}{4}\qquad s=\textit{length of one side}\\\\
-----------------------------\\\\
\textit{area of a circle}\\\\
A=\pi r^2\qquad r=radius\\\\
\textit{area of half a circle}\\\\
A=\cfrac{\pi r^2}{2}\\\\
-----------------------------\\\\$

$\bf \textit{now, let us use the side of 6, and radius of 3}
\\\\\\

\begin{array}{clclll}
\cfrac{6^2\sqrt{3}}{4}&-&\cfrac{\pi 3^2}{2}\\
\uparrow &&\uparrow \\
triangle's&&semi-circle's
\end{array}\impliedby \textit{area of shaded area}\\\\
-----------------------------\\\\
\boxed{\cfrac{36\sqrt{3}}{4}-\cfrac{9\pi }{2}}$

you can, add the fractions if you want, or leave them like that, or get their difference by using their decimal format