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4 years ago

AB and BC form a right angle at their point of intersection, B.

Mathematics

1 answer:

const2013 [10]4 years ago

8 0

Answer:

The y-intercept of AB is (4/3) and the equation of BC is y = (-1/6) x + (4/3)

If the y-coordinate of point C is 13, its x-coordinate is 4.

=====================================

Step-by-step explanation:

AB and BC form a right angle at their point of intersection, B.

So, AB ⊥ BC

A = (14 , -1) , B = (2,1)

Part A: Find the y-intercept of AB.

The general equation of the line y = mx + c

Where: m is the slope and c is constant represents y-intercept.

m = (y₂ - y₁)/(x₂ - x₁)

m = $\frac{1 - (-1)}{2-14} = \frac{2}{-12} = - \frac{1}{6}$

y = (-1/6) x + c

Substitute with the point (2,1) to find c

1 = (-1/6) * 2 + c

c = 1 + 2 *(1/6) = 4/3

y = (-1/6) x + 4/3

So, the y-intercept of AB is 4/3

=======================================

Part B: Find the equation of BC.

BC ⊥ AB at point B

if m is the slope of AB, the slope of BC = (-1/m)

m = (-1/6)

So, the slope of BC = (-1/m) = 6

The equation of the line BC ⇒ y = 6x + c

Substitute with the point (2,1) to find c

1 = 6 * 2 + c

c = 1 - 12 = -11

y = 6x - 11

=========================================

Part C: Find x-coordinate of C if y-coordinate of point C is 13.

The equation of the BC is y = 6x - 11

Substitute with the y-coordinate of point C = 13

13 = 6x - 11

6x = 13 + 11 = 24

x = 24/6 = 4

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The graph shown below expresses a radical function that can be written in the form f(x)= a(x + k)^1/n +C. What does the graph te

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Answer:

C. c is less than zero

Step-by-step explanation:

The parent radical function y=x^(1/n) has its point of inflection at the origin. The graph shows that point of inflection has been translated left and down.

<h3>Function transformation</h3>

The transformation of the parent function y=x^(1/n) into the function ...

f(x) = a(x +k)^(1/n) +c

represents the following transformations:

vertical scaling by a factor of 'a'
left shift by k units
up shift by c units

<h3>Application</h3>

The location of the inflection point at (-3, -4) indicates it has been shifted left 3 units, and down 4 units. In the transformed function equation, this means ...

k = 3
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2 years ago

An examination of the records for a random sample of 16 motor vehicles in a large fleet resulted in the sample mean operating co

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Answer:

1. The 95% confidence interval would be given by (24.8190;27.8010)

2. $4.7048 \leq \sigma^2 \leq 16.1961$

3. $t=\frac{26.31-25}{\frac{2.8}{\sqrt{16}}}=1.871$

$t_{crit}=1.753$

Since our calculated value it's higher than the critical value we have enough evidence at 5% of significance to reject th null hypothesis and conclude that the true mean is higher than 25 cents per mile.

4. $t=(16-1) [\frac{2.8}{2.3}]^2 =22.2306$

$\chi^2 =24.9958$

Since our calculated value is less than the critical value we don't hav enough evidence to reject the null hypothesis at the significance level provided.

Step-by-step explanation:

Previous concepts

$\bar X=26.31$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=2.8 represent the sample standard deviation

n=16 represent the sample size

Part 1

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

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=16-1=15$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.025,15)".And we see that $t_{\alpha/2}=2.13$

Now we have everything in order to replace into formula (1):

$26.31-2.13\frac{2.8}{\sqrt{16}}=24.819$

$26.31+2.13\frac{2.8}{\sqrt{16}}=27.801$

So on this case the 95% confidence interval would be given by (24.8190;27.8010)

Part 2

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

$df=n-1=16-1=15$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical values.

The excel commands would be: "=CHISQ.INV(0.05,15)" "=CHISQ.INV(0.95,15)". so for this case the critical values are:

$\chi^2_{\alpha/2}=24.996$

$\chi^2_{1- \alpha/2}=7.261$

And replacing into the formula for the interval we got:

$\frac{(15)(2.8)^2}{24.996} \leq \sigma^2 \leq \frac{(15)(2.8)^2}{7.261}$

$4.7048 \leq \sigma^2 \leq 16.1961$

Part 3

We need to conduct a hypothesis in order to determine if actual mean operating cost is at most 25 cents per mile , the system of hypothesis would be:

Null hypothesis: $\mu \leq 25$

Alternative hypothesis: $\mu > 25$

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{26.31-25}{\frac{2.8}{\sqrt{16}}}=1.871$

Critical value

On this case we need a critical value on th t distribution with 15 degrees of freedom that accumulates 0.05 of th area on the right and 0.95 of the area on the left. We can calculate this value with the following excel code:"=T.INV(0.95,15)" and we got $t_{crit}=1.753$

Conclusion

Since our calculated valu it's higher than the critical value we have enough evidence at 5% of significance to reject th null hypothesis and conclude that the true mean is higher than 25 cents per mile.

Part 4

State the null and alternative hypothesis

On this case we want to check if the population standard deviation is more than 2.3, so the system of hypothesis are:

H0: $\sigma \leq 2.3$

H1: $\sigma >2.3$

In order to check the hypothesis we need to calculate the statistic given by the following formula:

$t=(n-1) [\frac{s}{\sigma_o}]^2$

This statistic have a Chi Square distribution distribution with n-1 degrees of freedom.

What is the value of your test statistic?

Now we have everything to replace into the formula for the statistic and we got:

$t=(16-1) [\frac{2.8}{2.3}]^2 =22.2306$

What is the critical value for the test statistic at an α = 0.05 significance level?

Since is a right tailed test the critical zone it's on the right tail of the distribution. On this case we need a quantile on the chi square distribution with 15 degrees of freedom that accumulates 0.05 of the area on the right tail and 0.95 on the left tail.

We can calculate the critical value in excel with the following code: "=CHISQ.INV(0.95,15)". And our critical value would be $\chi^2 =24.9958$

Since our calculated value is less than the critical value we FAIL to reject the null hypothesis.

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