Answer: D) 
Step-by-step explanation:
Based on the provided information, we have
Sample size: n= 15
sample mean: 
Sample standard deviation: s= $20
Since the population standard deviation is not known, we utilize a t-test.
For a significance level of 95% confidence: 
Critical t-value :
[Using the Student's t-value table]
The required 95% confidence interval yields:-

Thus, the sought-after 95% confidence interval for the mean amount spent by credit card customers during their initial visit to the new store in the mall, assuming normal distribution of the spending amounts, is:

First note the ratio is
7:2 so divide 7 by 2 to get 7/2 = 3.5. Then add 3.5 to 2, yielding the result 5.5
For one day:
= £9.20 × 7
= £64.40
For six days:
£64.40 × 6 = £386.40
After sharing with his mom:
£386.40/7 × 5
= £55.20 × 5
= £276
To purchase a car worth £1932:
£1932/£276 = 7
Thus, he needs 7 weeks to save enough for the car priced at £1932.
To calculate, simply multiply 12 by 0.15 or 15%, which equals 1.8. Then subtract 1.8 from 12, yielding $10.20 as the discounted price for the pizza.
I hope this is helpful