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4 years ago

7

What is the slope of the line perpendicular to the line with the equation 5x + 3y=18?

1 answer:

Allushta [10]4 years ago

3 0

Answer:

3 i think

hope it helps...

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Rewrite in simplest form (-4x+6y-)-(-9x-8y

Alex Ar [27]

Answer:

5x + 14y

Explanation:

−4x+6y−(−9x−8y)

Distribute the Negative Sign:

= −4x+6y+−1(−9x−8y)

= −4x+6y+−1(−9x)+−1(−8y)

= −4x+6y+9x+8y

Combine Like Terms:

= −4x+6y+9x+8y

= (−4x+9x)+(6y+8y)

= 5x + 14y

5 0

3 years ago

What fraction of £1 is seventy pence?

Veronika [31]

The answer is: "7/10" .
________________________
Note: £1 = 100 pence

70/100 = (70÷10)/(100÷10) = 7/10 .
____________________________

4 0

4 years ago

Mr George has $59.75 in his bank account, but then he bought some new ties and some pancakes. After the purchase, he was in debt

Bezzdna [24]

-15.35 is the answer I think

5 0

4 years ago

Please help! I don't understand how to solve this problem

Ilia_Sergeevich [38]

Using the z-distribution, a sample of 142,282 should be taken, which is not practical as it is too large of a sample.

<h3>What is a z-distribution confidence interval?</h3>

The confidence interval is:

$\overline{x} \pm z\frac{\sigma}{\sqrt{n}}$

The margin of error is:

$M = z\frac{\sigma}{\sqrt{n}}$

In which:

$\overline{x}$ is the sample mean.

z is the critical value.

n is the sample size.

$\sigma$ is the standard deviation for the population.

Assuming an uniform distribution, the standard deviation is given by:

$S = \sqrt{\frac{(4 - 0)^2}{12}} = 1.1547$

In this problem, we have a 95% confidence level, hence $\alpha = 0.95$ , z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$ , so the critical value is z = 1.96.

The sample size is found solving for n when the margin of error is of M = 0.006, hence:

$M = z\frac{\sigma}{\sqrt{n}}$

$0.006 = 1.96\frac{1.1547}{\sqrt{n}}$

$0.006\sqrt{n} = 1.96 \times 1.1547$

$\sqrt{n} = \frac{1.96 \times 1.1547}{0.006}$

$(\sqrt{n})^2 = \left(\frac{1.96 \times 1.1547}{0.006}\right)^2$

n = 142,282.

A sample of 142,282 should be taken, which is not practical as it is too large of a sample.

More can be learned about the z-distribution at brainly.com/question/25890103

#SPJ1

8 0

1 year ago

Prove that: (secA-cosec A) (1+cot A +tan A) =( sec^2A/cosecA)-(Cosec^2A/secA)<br>

Ksju [112]

Step-by-step explanation:

$(\sec A - \csc A)(1 + \cot A + \tan A)$

$=(\sec A - \csc A)\left(1 + \dfrac{\cos A}{\sin A} + \dfrac{\sin A}{\cos A} \right)$

$=(\sec A - \csc A)\left(1 + \dfrac{\cos^2 A + \sin^2 A}{\sin A\cos A} \right)$

$=(\sec A - \csc A)\left(\dfrac{1 + \sin A \cos A}{\sin A \cos A} \right)$

$=\left(\dfrac{\frac{1}{\cos A} - \frac{1}{\sin A}+\sin A - \cos A}{\sin A\cos A}\right)$

$=\dfrac{\sin A - \sin A \cos^2A - \cos A + \cos A\sin^2A}{(\sin A\cos A)^2}$

$=\dfrac{\sin A(1 - \cos^2A) - \cos A (1 - \sin^2 A)}{(\sin A\cos A)^2}$

$=\dfrac{\sin^3A - \cos^3A}{\sin^2A\cos^2A}$

$=\dfrac{\sin A}{\cos^2A} - \dfrac{\cos A}{\sin^2A}$

$=\left(\dfrac{1}{\cos A}\right)\left(\dfrac{\sin A}{1}\right) - \left(\dfrac{1}{\sin^2A}\right) \left(\dfrac{\cos A}{1}\right)$

$=\sec^2A\csc A - \csc^2A\sec A$

5 0

3 years ago