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ELEN [110]
3 years ago
15

Write the equation of the line that passes through the given points (0,1) (-7,-5)

Mathematics
2 answers:
pishuonlain [190]3 years ago
6 0

Answer:

Step-by-step explanation:

andre [41]3 years ago
4 0

<u>ANSWER:  </u>

The line equation that passes through the given points (0,1) (-7,-5) is 6x – 7y + 7 = 0.

<u>SOLUTION: </u>

Given, two points are A(0, 1) and B(-7, -5).

We need to find the line equation that passes through the given two points.

We know that, general equation of a line passing through two points $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right),\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)$ is given by

$y-y_{1}=\left(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\right)\left(x-x_{1}\right)$ --- 1

Here,in our problem \mathrm{x}_{1}=0, \mathrm{y}_{1}=1, \mathrm{x}_{2}=-7$ and $\mathrm{y}_{2}=-5$

Now substitute the values in (1)

$y-1=\left(\frac{-5-1}{-7-0}\right)(x-0)$

$y-1=\frac{-6}{-7}(x)$

$y-1=\frac{6}{7} x$

7y – 7 = 6x

6x – 7y + 7 = 0

Hence, the line equation that passes through the given points is 6x – 7y + 7 = 0.

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Write an equation of the line using the given information:
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How many times must we toss a coin to ensure that a 0.95-confidence interval for the probability of heads on a single toss has l
musickatia [10]

Answer:

(1) 97

(2) 385

(3) 9604

Step-by-step explanation:

The (1 - <em>α</em>) % confidence interval for population proportion is:

CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}

The margin of error in this interval is:

MOE= z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}

The formula to compute the sample size is:

\\n=\frac{z_{\alpha/2}^{2}\times \hat p(1-\hat p)}{MOE^{2}}

(1)

Given:

\hat p = 0.50\\MOE=0.1\\z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

*Use the <em>z</em>-table for the critical value.

Compute the value of <em>n</em> as follows:

\\n=\frac{z_{\alpha/2}^{2}\times \hat p(1-\hat p)}{MOE^{2}}\\=\frac{1.96^{2}\times0.50\times(1-0.50)}{0.1^{2}}\\=96.04\\\approx97

Thus, the minimum sample size required is 97.

(2)

Given:

\hat p = 0.50\\MOE=0.05\\z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

*Use the <em>z</em>-table for the critical value.

Compute the value of <em>n</em> as follows:

\\n=\frac{z_{\alpha/2}^{2}\times \hat p(1-\hat p)}{MOE^{2}}\\=\frac{1.96^{2}\times0.50\times(1-0.50)}{0.05^{2}}\\=384.16\\\approx385

Thus, the minimum sample size required is 385.

(3)

Given:

\hat p = 0.50\\MOE=0.01\\z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

*Use the <em>z</em>-table for the critical value.

Compute the value of <em>n</em> as follows:

\\n=\frac{z_{\alpha/2}^{2}\times \hat p(1-\hat p)}{MOE^{2}}\\=\frac{1.96^{2}\times0.50\times(1-0.50)}{0.01^{2}}\\=9604

Thus, the minimum sample size required is 9604.

8 0
3 years ago
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