Simplify: -3(m – 2) – 7m A. -10m + 6 B. -4m + 6 C. 4m – 6 D. 10m + 6

Mathematics

2 answers:

Sholpan [36]3 years ago

8 0

<h2>Answer:The answer is the letter A</h2>

Masja [62]3 years ago

4 0

The correct answer is A

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Use the formula C(n, x)pxqn − x to determine the probability of the given event. (Round your answer to four decimal places.) The

miskamm [114]

Answer: 0.1334

Step-by-step explanation:

Given

There are seven trails with a probability of success is $p=\frac{1}{4}$

$\therefore q=1-p\\q=1-\frac{1}{4}\\q=\frac{3}{4}$

Using the binomial distribution, the Probability of exactly no success i.e. x=0 is given by

$\Rightarrow ^7C_0\cdot p^0 \cdot q^{7-0}\\\Rightarrow 1\times(\frac{1}{4})^0 \cdot (\frac{3}{4})^7\\\Rightarrow 1\times 1 \cdot (0.75)^7\\\Rightarrow 0.1334$

7 0

3 years ago

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.

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

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