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SOVA2 [1]
4 years ago
6

Which expression is equivalent to 5(x - 3y + 4)

Mathematics
2 answers:
defon4 years ago
4 0

Answer:

D. 5x-15y+20

Step-by-step explanation:

If you distribute 5 to all the numbers in the parenthesis, you get 5x-15y+20. This is because 5 multiplied by x is 5x, 5 multiplied by 3y is 15y, and 5 multiplied by 4 is 20. I hope this helps!

valentinak56 [21]4 years ago
4 0

Answer:

D

Step-by-step explanation:

you have to distribute the number 5 between all of the numbers. Is because

5 * x is 5x  3y is 15y, and 5 *4 is 20.

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3 years ago
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.

5 0
3 years ago
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