Solution :
The null and alternative hypothesis is given by
Assume that the level of significance, α = 0.05
The t-test statistics is, t = 1.484
Degree of freedom :
df = n - 1
= 14 - 1
= 13
The P-value is given by
P-value = 2P (T>|t|)
= 2P(T>|1.484|)
= 2P(T>1.484)
= 2(=T.DIST.RT(1.484,13))
= 0.05
7/5 * 5/7y = 7/5 * 6
y = 42/5.
The multiplication property of equality is used.
Answer:
3.6 repeating or 3 and 2/3
Step-by-step explanation:
I did the work on a calculator.
Integer 1 = n
integer 2 = n + 1
integer 3 = n + 2
n + (n + 1) + (n + 2) = 141
3n + 3 = 141
3n = 141 - 3
3n = 138
n = 138/3
n = 46 --(1st integer)
n + 1 = 46 + 1 = 47 --(2nd integer)
n + 2 = 46 + 2 = 48 --(3rd integer)
so the smallest integer is 46
Draw whats on the paper, thats your answer
