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vazorg [7]
3 years ago
14

• What is the solution to 5x+ 20 = 7(x+2)?

Mathematics
1 answer:
o-na [289]3 years ago
6 0

For this case we must find the solution of the following equation:

5x + 20 = 7 (x + 2)

Applying distributive property on the right side of the equation we have:

5x + 20 = 7x + 14

Subtracting 7x from both sides of the equation:

5x-7x + 20 = 14\\-2x + 20 = 14

Subtracting 20 from both sides of the equation:

-2x = 14-20

Different signs are subtracted and the major sign is placed.

-2x = -6

Dividing between -2 on both sides of the equation:

x = \frac {-6} {- 2}\\x = 3

Thus, the solution of the equation isx = 3

Answer:

x = 3

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The confidence Interval is [- 0.7053  10.4521]

a: The  hypotheses  are

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

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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

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s1= 5.6       s2= 4.3

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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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