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3 years ago

8

Identify an equation in point-slope form for the line perpendicular to y=-4x-1 that passes through (-2,7)

2 answers:

sesenic [268]3 years ago

7 0

Answer:

Y-7=1/4(x+2)

Step-by-step explanation:

stiks02 [169]3 years ago

4 0

The slope perpendicular to $y=-4x-1$ is the negative reciprocal, or $\frac{1}{4}$ .
We want it to pass through the point (-2,7).

$y-7= \frac{1}{4}[x-(-2)]$

Hope this helps! :^)

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Final grades are often computed using weighted averages. For example, Fran's teacher uses the following: homework 10%, quizzes 2

STALIN [3.7K]

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Final grade = (0.10 x 92) + ( 0.20 x 68) + (0.70 x 81)

<span>Final grade = 79.5 </span>

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3 years ago

Answer asap thank you!??!!??!

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2 years ago

A model of a plane is made using a scale of 1:5. If the wings of the model are 1.2m long, what is the actual length of the wings

7nadin3 [17]

Answer:

6 m

Step-by-step explanation:

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3 years ago

Factorise the following:<br>1. 3p - 3q<br>2. 8m + 12n + 16 r<br>3. 4x² - 6xy

UNO [17]

Step-by-step explanation:

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3 years ago

The diameters of aluminum alloy rods produced on an extrusion machine are known to have a standard deviation of 0.0001 in. A ran

Murljashka [212]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The null hypothesis is rejected this means that $\mu \ne 0.5025$

The 95% confidence interval is $0.5045608 < \mu < 0.5046392$

Step-by-step explanation:

From the question we are told that

The standard deviation is s= 0.0001

The sample size is n = 25

The sample mean is $\= x = 0.5046 \ in$

The population mean is $\mu = 0.5025 \ in$

The null hypothesis is $H_o : \mu = 0.5025$

The alternative hypothesis is $H_a : \mu \ne 0.5025$

The test statistics is mathematically represented as

$t = \frac{0.5046 - 0.5025}{ \frac{0.0001}{\sqrt{25} } }$

$t = 105$

So the p-value from the z-table is mathematically represented as

$p-value = 2 * P( z > 105)$

$p-value = 0.000$

seeing that

$p-value < \alpha$ we reject the null hypothesis

The critical value of

$\frac{\alpha }{2}$ obtained from the normal distribution table is

$Z_{\frac{\alpha }{2} } = 1.96$

The margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} }*\frac{s}{\sqrt{n} }$

=> $E = 1.96 *\frac{0.0001}{\sqrt{25} }$

=> $E =3.92 *10^{-5}$

The 95% confidence level is mathematically represented as

$\= x - E < \mu < \= x + E$

=> $0.5046 - 3.92 *10^{-5} < \mu < 0.5046 + 3.92 *10^{-5}$

=> $0.5045608 < \mu < 0.5046392$

8 0

3 years ago