Consider two populations of coins, one of the pennies and one of quarters. A random sample of 25 pennies was selected, and the mean age of the sample was 32 years. A random sample of 35 quarters was taken, and the mean age of the sample was 19 years.
For the sampling distribution of the difference in sample means, have the conditions for normality been met?

a. Yes, the conditions for normality have been met because the distributions of age for the two populations are approximately normal.
b. Yes, the conditions for normality have been met because the sample sizes taken from both populations are large enough.
c. No, the conditions for normality have not been met because neither sample size is large enough and no information is given about the distributions of the populations.
d. No, the conditions for normality have not been met because the sample size for the pennies is not large enough and no information is given about the distributions of the populations.
e. No, the conditions for normality have not been met because the sample size for the quarters is not large enough and no information is given about the distributions of the populations.

d. No, the conditions for normality have not been met because the sample size for the pennies is not large enough and no information is given about the distributions of the populations.

Step-by-step explanation:

In this example, we learn that the sample taken of pennies was of 25, while the sample of quarters was of 35. According to the central limit theorem, a sample must be equal to or larger than 30 for the theorem to hold. The theorem states that when random variables are added, they tend towards a normal distribution even if the original variables themselves are not normally distributed. In this question, the sample of pennies is not large enough for the theorem to hold. On top of this, we have no information about the distribution of the populations. Therefore, the conditions for normality have not been met.

a.   we have two

lines:   y = 4-x   and  

y = 8-x^-1

given two simultaneous equations that are both to be

true, then the solution is the points where the lines cross. the intersection is

where the two equations are equal. therefore the solution that works for both

equations is when

4-x = 8-x^-1

this is where the two lines will cross and that is the

common point that satisfies both equations.

b.   4-x = 8-x^-1

  x           4-x       8-x^-1

-3           7        

8.33

-2           6         8.5

-1           5         9

  0           4        

-

  1           3        

7

  2           2        

7.5

  3           1         7.67

the table shows that none of the x values from -3 to 3 is

the solution because in no case does

4-x = 8-x^-1

to find the solution we need to rearrange the equation to

find for x:

4-x = 8-x^-1

multiply both sides with x:

4x-x^2 = 8x-1

x^2+4x-1=0

x= -4.236, 0.236

therefore there are two points that satisfies the

equation.

find y:  

x=-4.236

y = 4-x   = 4 – (-4.236)

= 8.236

y = 8-x^-1 =   .236)^-1

= 8.236

x=0.236

  y = 4-x   = 4 – (0.236) = 3.764

y = 8-x^-1 =   8-(0.236)^-1

= 3.764

thus the two lines cross at 2 points:

(-4.236, 8.236) & (0.236, 3.764)

c.   to solve

graphically the equation 4-x = 8-x^-1

we would graph both lines: y = 4-x   and   y

= 8-x^-1

the point on the graph where the lines cross is the

solution to the system of equations.

just graph the points on part b on a cartesian coordinate

system and extend the two lines.   the

solution is, as stated, the point where the two lines cross on the graph.

step-by-step explanation:

mark brainliest!