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3 years ago

What is the solution to the following equation? 5(2x –6) + 20 = 10

Mathematics

2 answers:

Igoryamba3 years ago

3 0

Answer:

Step-by-step explanation:

Simplifying

5(2x + -6) + 20 = 10

Reorder the terms:

5(-6 + 2x) + 20 = 10

(-6 * 5 + 2x * 5) + 20 = 10

(-30 + 10x) + 20 = 10

Reorder the terms:

-30 + 20 + 10x = 10

Combine like terms: -30 + 20 = -10

-10 + 10x = 10

Solving

-10 + 10x = 10

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '10' to each side of the equation.

-10 + 10 + 10x = 10 + 10

Combine like terms: -10 + 10 = 0

0 + 10x = 10 + 10

10x = 10 + 10

Combine like terms: 10 + 10 = 20

10x = 20

Divide each side by '10'.

x = 2

Simplifying

x = 2

OlgaM077 [116]3 years ago

3 0

Answer:

x=2

Step-by-step explanation:

Let's solve your equation step-by-step.

5(2x−6)+20=10

Step 1: Simplify both sides of the equation.

5(2x−6)+20=10

(5)(2x)+(5)(−6)+20=10(Distribute)

10x+−30+20=10

(10x)+(−30+20)=10(Combine Like Terms)

10x+−10=10

10x−10=10

Step 2: Add 10 to both sides.

10x−10+10=10+10

10x=20

Step 3: Divide both sides by 10.

10x

x=2

Answer:

x=2

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Rudik [331]

Answer:

a) $0.93 - 1.96\sqrt{\frac{0.93(1-0.93)}{500}}=0.908$

$0.93 + 1.96\sqrt{\frac{0.93(1-0.93)}{500}}=0.952$

The 95% confidence interval would be given by (0.908;0.0.952)

b) $0.21 - 2.58\sqrt{\frac{0.21(1-0.21)}{500}}=0.163$

$0.21 + 2.58\sqrt{\frac{0.21(1-0.21)}{500}}=0.257$

The 99% confidence interval would be given by (0.163;0.0.257)

c) The margin of error for part a is:

$ME= 1.96\sqrt{\frac{0.93(1-0.93)}{500}}=0.0224$

And for part b is:

$ME=2.58\sqrt{\frac{0.21(1-0.21)}{500}}=0.0470$

So then the margin of error is larger for part b.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Part a

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

The confidence interval for the mean is given by the following formula:

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

If we replace the values obtained we got:

$0.93 - 1.96\sqrt{\frac{0.93(1-0.93)}{500}}=0.908$

$0.93 + 1.96\sqrt{\frac{0.93(1-0.93)}{500}}=0.952$

The 95% confidence interval would be given by (0.908;0.0.952)

Part b

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$ . And the critical value would be given by:

$z_{\alpha/2}=-2.58, z_{1-\alpha/2}=2.58$

The confidence interval for the mean is given by the following formula:

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

If we replace the values obtained we got:

$0.21 - 2.58\sqrt{\frac{0.21(1-0.21)}{500}}=0.163$

$0.21 + 2.58\sqrt{\frac{0.21(1-0.21)}{500}}=0.257$

The 99% confidence interval would be given by (0.163;0.0.257)

Part c

The margin of error for part a is:

$ME= 1.96\sqrt{\frac{0.93(1-0.93)}{500}}=0.0224$

And for part b is:

$ME=2.58\sqrt{\frac{0.21(1-0.21)}{500}}=0.0470$

So then the margin of error is larger for part b.

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