Answer:
Kindly check explanation
Step-by-step explanation:
The hypothesis :
H0 : μ = 3000
H0 : μ ≠ 3000
The test statistic :
(xbar - μ) ÷ (s/√(n))
xbar = 3500
μ = 3000
σ = 300
n = 30
(3500 - 3000) ÷ (350/√(30))
Test statistic = 7.824
Df = 30 - 1 = 29
Tcritical at 0.01 = 2.462
Test statistic > critical value ; we reject H0 ; and concluded that there is significant evidence that
μ ≠ 3000
360 hope that this helped
Answer:
365.4
Step-by-step explanation:
i dont know what £ means..... sorry i just did 9x40.60
I think that the sum will always be a rational number
let's prove that
<span>any rational number can be represented as a/b where a and b are integers and b≠0
</span>and an integer is the counting numbers plus their negatives and 0
so like -4,-3,-2,-1,0,1,2,3,4....
<span>so, 2 rational numbers can be represented as
</span>a/b and c/d (where a,b,c,d are all integers and b≠0 and d≠0)
their sum is
a/b+c/d=
ad/bd+bc/bd=
(ad+bc)/bd
1. the numerator and denominator will be integers
2. that the denominator does not equal 0
alright
1.
we started with that they are all integers
ab+bc=?
if we multiply any 2 integers, we get an integer
<span>like 3*4=12 or -3*4=-12 or -3*-4=12, etc.
</span>even 0*4=0, that's an integer
the sum of any 2 integers is an integer
like 4+3=7, 3+(-4)=-1, 3+0=3, etc.
so we have established that the numerator is an integer
now the denominator
that is just a product of 2 integers so it is an integer
<span>2. we originally defined that b≠0 and d≠0 so we're good
</span>therefore, the sum of any 2 rational numbers will always be a rational number <span>is the correct answer.</span>
