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3 years ago

Find the distance and midpoint for (2, 0, -2) and (5, -4, 6)

1 answer:

7 0

Hi, It is well simples:

The distance between two points is :

d( A, B) = √[ (xb - xa)^2+(yb - ya)^2+(zb - za)^2]

Then,

A = (xa, ya, za )

B = (xb , yb , zb)

We know:

A = ( 2 ,0 ,-2)

And,

B = (5 , -4 , 6)
___________

xb - xa = 5-2 <=> 3

yb - ya = -4 - 0 <=> -4

zb - za = 6 - (-2) <=> 8

Then us stay:

d( A, B) = √[ (3)^2 + (-4)^2+(8)^2]

= √( 9 + 16 + 64)

= √( 25 + 64)

= √(89)

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The coordinates of the vertices of a quadrilateral are (8, 11) (11, 7) (8, 3) (2, 3). What is the area, in square units, of the

disa [49]

Check the picture below.

so the quadrilateral is really just a parallelogram below a triangle, so let's simply get the area of each and sum them up.

$\stackrel{\textit{\Large Areas}}{\stackrel{parallelogram}{(6)(4)}~~ + ~~\stackrel{triangle}{\cfrac{1}{2}(6)(4)}}\implies 24~~ + ~~12\implies 36$

6 0

2 years ago

What is 2.9% of 1000

777dan777 [17]

' 2.9% ' means ' 0.029 '

' of ' means ' times '

' 2.9% of 1,000 ' means ' 0.029 x 1,000 '

0.029 x 1,000 = 29

3 0

3 years ago

Read 2 more answers

Randomly selected 110 student cars have ages with a mean of 8 years and a standard deviation of 3.6 years, while randomly select

monitta

Answer:

1. Yes, there is sufficient evidence to support the claim that student cars are older than faculty cars.

2. The 98% confidence interval for the difference between the two population means is [1.432 years, 3.968 years].

Step-by-step explanation:

We are given that randomly selected 110 student cars to have ages with a mean of 8 years and a standard deviation of 3.6 years, while randomly selected 75 faculty cars to have ages with a mean of 5.3 years and a standard deviation of 3.7 years.

Let $\mu_1$ = mean age of student cars.

$\mu_2$ = mean age of faculty cars.

So, Null Hypothesis, $H_0$ : $\mu_1 \leq \mu_2$ {means that the student cars are younger than or equal to faculty cars}

Alternate Hypothesis, $H_A$ : $\mu_1>\mu_2$ {means that the student cars are older than faculty cars}

(1) The test statistics that will be used here is Two-sample t-test statistics because we don't know about the population standard deviations;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)} {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t_n_1_+_n_2_-_2$

where, $\bar X_1$ = sample mean age of student cars = 8 years

$\bar X_2$ = sample mean age of faculty cars = 5.3 years

$s_1$ = sample standard deviation of student cars = 3.6 years

$s_2$ = sample standard deviation of student cars = 3.7 years

$n_1$ = sample of student cars = 110

$n_2$ = sample of faculty cars = 75

Also, $s_p=\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(110-1)\times 3.6^{2}+(75-1)\times 3.7^{2} }{110+75-2} }$ = 3.641

So, the test statistics = $\frac{(8-5.3)-(0)} {3.641 \times \sqrt{\frac{1}{110}+\frac{1}{75} } }$ ~ $t_1_8_3$

= 4.952

The value of t-test statistics is 4.952.

Since the value of our test statistics is more than the critical value of t, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we support the claim that student cars are older than faculty cars.

(2) The 98% confidence interval for the difference between the two population means ( $\mu_1-\mu_2$ ) is given by;

98% C.I. for ( $\mu_1-\mu_2$ ) = $(\bar X_1-\bar X_2) \pm (t_(_\frac{\alpha}{2}_) \times s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} })$

= $(8-5.3) \pm (2.326 \times 3.641 \times \sqrt{\frac{1}{110}+\frac{1}{75} })$

= $[2.7 \pm 1.268]$

= [1.432, 3.968]

Here, the critical value of t at a 1% level of significance is 2.326.

Hence, the 98% confidence interval for the difference between the two population means is [1.432 years, 3.968 years].

7 0

3 years ago

How long will it take a man to walk 7 km if he has walked 4 km for one hour?

Semmy [17]

See the photo mate.As we know time=speed/time than as he travelled 4km in 1 hour probably his speed is 4km/h than time required to travell 7km

Time=7/4=1.75hours

3 0

3 years ago

The length, in centimeters, of a diagonal of a rectangle is represented by the expression below.

TiliK225 [7]

Answer:

(C) 18 centimeters

4 0

2 years ago

Read 2 more answers