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Eva8 [605]
3 years ago
10

Given parallelogram ABCD with C(5,4), find the coordinates of A if the diagonals AC and BD intersect at (2,7).

Mathematics
2 answers:
finlep [7]3 years ago
8 0
Plot graph and count for A. 

A (-1,10)
Anna71 [15]3 years ago
4 0

Answer:

<h2>A (-1;10)</h2>

Step-by-step explanation:

From the problem, we deduct that point (2,7) is a middle point for diagonals AC and BD, because the intersection of diagonals in a parallelogram intercept in the middle point of each one.

So, the coordinates of A are (x,y) unknown, and coordinates of C are (5,4), applying the definition of middle point:

m_{AC}=(\frac{x+5}{2};\frac{y+4}{2} )

However we know the exact coordinates of the middle point, which are (2,7), that means:

\frac{x+5}{2}=2\\ x=4-5=-1

\frac{y+4}{2}=7\\ y=14-4=10

Therefore, the coordinates of A are (-1;10)

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Find the exact value of sin 5pi/4
AURORKA [14]

The exact value of sin 5π/4 is 0. 0677

<h3>How to find the value</h3>

It is important to note that π = 3. 142

Let's find sin 5π

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The local branch of the Internal Revenue Service spent an average of 21 minutes helping each of 10 people prepare their tax retu
AURORKA [14]

Answer:

The confidence Interval is [- 0.7053  10.4521]

a: The  hypotheses  are

H0: μ1=μ2    against the claim Ha :μ1≠μ2

b. The critical value for t∝/2 for 17 d.f  t > 2.508 and  t < -2.111

c. t= -2.8422

d. The calculated value of t= -2.8422 is less than t < -2.11 the critical value therefore we reject H0 and conclude there is a difference between the two means.

Step-by-step explanation:

When the standard deviations are not the same then the confidence intervals for mean differences are calculated as

(x1`-x2`)- t∝/2 √s1²/n1 + s2²/n2 < u1-u2 < (x1`-x2`)+ t∝/2 √s1²/n1 + s2²/n2

x1`= 21        x2`= 27

n1=  10       n2= 14

s1= 5.6       s2= 4.3

The degrees of freedom is calculated using

υ = [s₁²/n1 + s₂²/n2]²/ (s₁²/n1 )²/ n1-1 + (s₂²/n2)²/n2-1

= 17

The t∝/2 for 17 d.f = 2.11

Putting the values

(x1`-x2`)- t∝/2 √s1²/n1 + s2²/n2 < u1-u2 < (x1`-x2`)+ t∝/2 √s1²/n1 + s2²/n2

(21-27) - 2.11√5.6²/10+ 4.3²/14 < u1-u2 <(21-27)  +2.11√5.6²/10+4.3²/14

6- 2.11*2.111 < u1-u2 <  ( 6 )  +2.11*2.111

6- 4.4521 < u1-u2 <  ( 6 )  +5.294

- 1.5479 < u1-u2 <  10.4521

The confidence Interval is [- 0.7053  10.4521]

a: The  hypotheses  are

H0: μ1=μ2    against the claim Ha :μ1≠μ2

The claim is that there is a difference in the average time spent by the two services

b. The critical value for t∝/2 for 17 d.f  t > 2.508 and  t < -2.111

The degrees of freedom is calculated using

υ = [s₁²/n1 + s₂²/n2]²/ (s₁²/n1 )²/ n1-1 + (s₂²/n2)²/n2-1

= 17

c. The test statistic is

t= (x1`-x2`)  /√s1²/n1 + s2²/n2

t= (21-27)  /√5.6²/10+ 4.3²/14

t= -6/2.111

t= -2.8422

d. The calculated value of t= -2.8422 is less than t < -2.11 the critical value therefore we reject H0 and conclude there is a difference between the two means.

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