A sample of 100 consumers shows that the average amount spent annually on a debit card is $8000. And assume that the population standard deviation of debit card expenditure is $500. Construct a 95% confidence interval of the population mean amount spent annually on a debit card.

The confidence interval is ($7902, $8098)

Step-by-step explanation:

The formula for confidence interval of a normal distribution is given as:

Confidence Interval = μ ± z × σ/√n

Where:

μ is the mean amount spent annually = $8,000

σ is the standard deviation = 500

n is the number of samples = 100 customers

z = 95% confidence interval = 1.96

Confidence Interval = $8000 ± 1.96 ×500√100

= $8000 ± 1.96 × 500/10

Confidence Interval = $8000 ± 1.96 × 50

= $8000 ± 98

= $8000 - 98 = $7902

= $8,000 + 98 = $8098

Therefore, the 95% confidence interval of the population mean amount spent annually on a debit card = ($7902, $8098)

7

step-by-step explanation:

yuo ned the activity or vertical angle of the form of 1,7 but 2,2 has a c4,2 as it verticie so i not have a car

x > 1.3

step by step:

8.5−0.9x> 9.8−1.9x

step 1: simplify both sides of the inequality.

−0.9x+8.5> −1.9x+9.8

step 2: add 1.9x to both sides.

−0.9x+8.5+1.9x> −1.9x+9.8+1.9x

1x+8.5> 9.8

step 3: subtract 8.5 from both sides.

1x+8.5−8.5> 9.8−8.5

1x> 1.3

step 4: divide both sides by 1.

1x/1 > 1.3/1