If a,b,c are the 3 positive integers
1/a +1/b +1/c > 6/abc
(bc+ac+ab)/abc >6/abc so
(bc+ac+ab)>6
The lowest positive integers that are different are 1,2,3 so the lowest value that (bc+ac+ab) could have is 1•2+2•3+1•3=2+6+3= 11 therefore
1/a +1/b +1/c > 6/abc is true
Answer:
128 inches
Step-by-step explanation:
X^2+16-38=0
find what 2 numbers multiply to get -38 and add to get 16
find factors of 38
2 times 19
there is no way to use common factor so use quadratic formula which is
x=

or x=

ax^2+bx+c=0 subsitte
a=1
b=16
c=-38
subsitute

=

=

=

=

so the answers for x are

and

which are aprox -2.099504938362 and +18.099504938362
so it would be
(x-[-8+√102])(x+[8+√102])=0
first you turn the fractions into inproper fractions by multiplying the whole number by the denomonator and adding the numerator
4×4=16
16+3=19
19/4
4×7=28
28+1=29
29/7
keep the denomonators the same. now that you've converted them into inproper fractions you can multiply them
19/4 × 29/7=551/28
there is no way to simplify this answer so this is the final answer
Answer: Option C. [0.95, 1.59].
Step-by-step explanation:
We know that the mean is:
M = 1.27
and the margin of error is:
e = 0.32
This means that the actual value can be at a maximum distance of 0.32 from the mean.
then the interval will be:
[M - e, M + e].= [1.27 - 0.32, 1.27 + 0.32].= [0.95, 1.59].
The correct option is C.