The number of times the people in a group surveyed on the street have gone to the cinema this month are:

1 answer:

3 0

I will help you do it not give the answer,
Ok so we draw a frequency table by like say if we have 3 2’s then we add a column of like 0-5 and say like 3 frequencies,
I know it’s confusing at first but just keep trying
And to find average we add all the numbers and we divide by how many numbers there are.
Hope it helps!
If it doesn’t I will explain again.
Can I get brainliest answer?

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A rectangle has a height of 5m^4 and a width of m^2-2m-1.

In-s [12.5K]

Answer:

area of the rectangle = 5(m⁶ - 2m⁵ - m⁴) or 5m⁴(m² - 2m - 1)

Step-by-step explanation:

The rectangle has a height of 5m⁴ and the width is m² - 2m - 1.The area of the rectangle is the product of the width and the height.

width = m² - 2m - 1

height = 5m⁴

area of rectangle = 5m⁴ × (m² - 2m - 1)

area of rectangle = 5m⁴(m² - 2m - 1)

area of the rectangle = 5m⁶ - 10m⁵ - 5m⁴

area of the rectangle = 5(m⁶ - 2m⁵ - m⁴)

The area of the entire rectangle can be expressed as 5(m⁶ - 2m⁵ - m⁴) or 5m⁴(m² - 2m - 1)

8 0

2 years ago

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Need pre-cal help. Will mark best answer brainliest

OlgaM077 [116]

so, let's keep in mind that

$\bf \begin{array}{|c|ll} \cline{1-1} \textit{a\% of b}\\ \cline{1-1} \\ \left( \cfrac{a}{100} \right)\cdot b \\\\ \cline{1-1} \end{array}$

so let's make a quick table of those solutions, say A, B, C solutions with x,y,z liters of acid, with an acidity of 0.25, 0.40 and 0.60 respectively.

$\bf \begin{array}{lcccl} &\stackrel{solution}{quantity}&\stackrel{\textit{\% of }}{amount}&\stackrel{\textit{liters of }}{amount}\\ \cline{2-4}&\\ A&x&0.25&0.25x\\ B&y&0.40&0.4y\\ C&z&0.60&0.6z\\ \cline{2-4}&\\ mixture&78&0.45&35.1 \end{array} \\\\\\ \begin{cases} x+y+z=78\\ 0.25x+0.4y+0.6z=35.1 \end{cases}$

we know she's using "z" liters and those are 3 times as much as "y" liters, so z = 3y.

$\bf \begin{cases} x+y+3y=78\\ x+4y=78\\[-0.5em] \hrulefill\\ 0.25x+0.4y+0.6(3y)=35.1\\ 0.25x+0.4y=1.8y=35.1\\ 0.25x+2.2y=35.1 \end{cases}\implies \begin{cases} x+4y=78\\\\ 0.25x+2.2y=35.1 \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ x+4y=78\implies \boxed{x}=78-4y \\\\\\ \stackrel{\textit{using substitution on the 2nd equation}}{0.25\left( \boxed{78-4y} \right)+2.2y=35.1}\implies 19.5-y+2.2y=35.1$

$\bf 1.2y=15.6\implies y=\cfrac{15.6}{1.2}\implies \blacktriangleright y=13 \blacktriangleleft \\\\\\ x=78-4y\implies x=78-4(13)\implies \blacktriangleright x=26 \blacktriangleleft \\\\\\ z=3y\implies z=3(13)\implies \blacktriangleright z=39 \blacktriangleleft \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{25\%}{26}\qquad \stackrel{40\%}{13}\qquad \stackrel{60\%}{39}~\hfill$