U can refer to the picture that I’ve attach below
7 % of cost price is 2.80 , What about 1%?
7 = 2.80
1 = x (where x is the price value of 1% in dollars)
We simplify the above equation, so:
7 × x = 2.8 × 1 → 7x = 2.8 → x = 2.8 / 7 → x = 0.4 dollars
We know that 100% represents the whole price,
so if 1% is 0.4 dollars, then 100% will be 0.4 × 100 = 40 dollars
Therefore the store charged $40 dollars for the helmet before tax
<span>280
I'm assuming that this question is badly formatted and that the actual number of appetizers is 7, the number of entres is 10, and that there's 4 choices of desserts. So let's take each course by itself.
You can choose 1 of 7 appetizers. So we have
n = 7
After that, you chose an entre, so the number of possible meals to this point is
n = 7 * 10 = 70
Finally, you finish off with a dessert, so the number of meals is:
n = 70 * 4 = 280
Therefore the number of possible meals you can have is 280.
Note: If the values of 77, 1010 and 44 aren't errors, but are actually correct, then the number of meals is
n = 77 * 1010 * 44 = 3421880
But I believe that it's highly unlikely that the numbers in this problem are correct. Just imagine the amount of time it would take for someone to read a menu with over a thousand entres in it. And working in that kitchen would be an absolute nightmare.</span>
Answer:
E) we will use t- distribution because is un-known,n<30
the confidence interval is (0.0338,0.0392)
Step-by-step explanation:
<u>Step:-1</u>
Given sample size is n = 23<30 mortgage institutions
The mean interest rate 'x' = 0.0365
The standard deviation 'S' = 0.0046
the degree of freedom = n-1 = 23-1=22
99% of confidence intervals
(from tabulated value).





using calculator

Confidence interval is


the mean value is lies between in this confidence interval
(0.0338,0.0392).
<u>Answer:-</u>
<u>using t- distribution because is unknown,n<30,and the interest rates are not normally distributed.</u>