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dezoksy [38]
3 years ago
7

Write an equation of the line that passes through the points (-2,-3) and (1,-3)

Mathematics
1 answer:
sladkih [1.3K]3 years ago
3 0

Answer:

y = -3

Step-by-step explanation:

-3--3/-2-1 = 0

so slope is 0

y = -3

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1/2x + 1/3y = 7
pashok25 [27]

let's multiply both sides in each equation by the LCD of all fractions in it, thus doing away with the denominator.

\begin{cases} \cfrac{1}{2}x+\cfrac{1}{3}y&=7\\\\ \cfrac{1}{4}x+\cfrac{2}{3}y&=6 \end{cases}\implies \begin{cases} \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{6}}{6\left( \cfrac{1}{2}x+\cfrac{1}{3}y \right)=6(7)}\\\\ \stackrel{\textit{multiplying both sides by }\stackrel{LCD}{12}}{12\left( \cfrac{1}{4}x+\cfrac{2}{3}y\right)=12(6)} \end{cases}\implies \begin{cases} 3x+2y=42\\ 3x+8y=72 \end{cases} \\\\[-0.35em] ~\dotfill

\bf \stackrel{\textit{using elimination}}{ \begin{array}{llll} 3x+2y=42&\times -1\implies &\begin{matrix} -3x \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~-2y=&-42\\ 3x+8y-72 &&~~\begin{matrix} 3x \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~+8y=&72\\ \cline{3-4}\\ &&~\hfill 6y=&30 \end{array}} \\\\\\ y=\cfrac{30}{6}\implies \blacktriangleright y=5 \blacktriangleleft \\\\[-0.35em] ~\dotfill

\bf \stackrel{\textit{substituting \underline{y} on the 1st equation}~\hfill }{3x+2(5)=42\implies 3x+10=42}\implies 3x=32 \\\\\\ x=\cfrac{32}{3}\implies \blacktriangleright x=10\frac{2}{3} \blacktriangleleft \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \left(10\frac{2}{3}~~,~~5 \right)~\hfill

7 0
4 years ago
What is 7/4 as a mixed number? A. 13/4 B. 11/4 C. 3/4 D. 17/4
natima [27]
7/4 as a mixed number is the mixed number 1 3/4
3 0
3 years ago
Read 2 more answers
Help me please!!!!!!!!!!!!!!!!!
Usimov [2.4K]

Answer:

69

Step-by-step explanation:

8 0
3 years ago
Suppose that a market research firm is hired to estimate the percent of adults living in a large city who have cell phones. 500
Alborosie

Answer:

The 95% confidence interval estimate for the true proportion of adults residents of this city who have cell phones is (0.81, 0.874).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

For this problem, we have that:

n = 500, \pi = \frac{421}{500} = 0.842

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.842 - 1.96\sqrt{\frac{0.842*0.158}{500}} = 0.81

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.842 + 1.96\sqrt{\frac{0.842*0.158}{500}} = 0.874

The 95% confidence interval estimate for the true proportion of adults residents of this city who have cell phones is (0.81, 0.874).

6 0
4 years ago
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