1.) The sum(addition) of 21 and 5 times(multiplication) a number f is(=) 61.
f = unknown number/variable [So 21 plus 5f(5 times f) equals 61]
21 + 5f = 61 [21(one-time) + 5f(number x variable) = 61(total)]
2.) Seventeen more(addition) than seven times(multiplication) a number j is(=) 87.
j = unknown number/variable [So 17 plus 7j(7 times j) equals 87]
17 + 7j = 87
3.) n = number of calls
18 + 0.05n = 50.50
[Company charges $18 plus five cents per call(n), and the total charge was $50.50]
4.) s = the number of students
40 + 30s = 220
[Tutor charges $40 plus $30 per student(s), and the total charge was $220]
Original price of the item = $14.95
Price after discount = $13.79
Discount offered = original price - price after discount = 14.95- 13.79 = $1.16
Now let us find the percentage of discount offered.
Percentage discount is given by the formula:
Where MP= Marked price= original price
SP= selling price= price after discount
Percentage discount = 7.759 %
Number of Neutrons = Atomic mass - Atomic number
= 16 - 7 = 9
In short, Your Answer would be 9
Hope this helps!
Answer:
Step-by-step explanation:
H0 : μ = 46300
H1 : μ > 46300
α = 0.05
df = n - 1 = 45 - 1 = 44
Critical value for a one tailed t-test(since population standard deviation is not given).
Tcritical = 1.30
The test statistic :(xbar - μ) ÷ (s/sqrt(n))
The test statistic, t= (47800-46300) ÷ (5600√45)
t = 1500
t = 1500 / 834.79871
t = 1.797
The decision region :
Reject H0: if t value > critical value
1. 797 > 1.30
Tvalue > critical value ; We reject H0
Hence, there is sufficient evidence to conclude that cost has increased.
