Hey!
To find the percentage increase, you must use this formula:
{(50 - 92 / 50] × 100 = 84%
So, the percent increase from 50 to 92 is: 84%.
Answer:
15.
Step-by-step explanation:
The new average will be 11 + 4 = 15.
Answer:
-1.23076923077
Step-by-step explanation:
Step-by-step explanation:
the introduction of a fraction tells us that we are dealing with multiplications, and therefore a geometric sequence (where every new term is created by multiplying the previous term by a constant factor, the ratio r).
I think your teacher made a mistake, or you made one when typing the question in here.
there is no factor r that creates
15×r = 9
and
9×r = 5/27
it would mean that
15 × r² = 5/27
r² = 5/27 / 15 = 5/27 × 1/15 = 5/405 = 1/81
r = 1/9
but 15 × 1/9 = 5 × 1/3 = 5/3 is NOT 9
and 9 × 1/9 = 9/9 = 1 is NOT 5/27
so, this can't be right.
on the other hand
15 × r = 9
r = 9/15 = 3/5
and then
9 × 3/5 = 27/5
so, either the sequence should have been
15, 5/3, 5/27
or (and I suspect this to be true)
15, 9, 27/5
under that assumption we have
s1 = 15
r = 3/5
sn = sn-1 × r = s1 × r^(n-1) = 15 × (3/5)^(n-1)
s10 = 15 × (3/5)⁹ = 15 × 19683/1953125 =
= 3 × 19683/390625 = 59049/390625 =
= 0.15116544 ≈ 0.151
Answer:
Type I: 1.9%, Type II: 1.6%
Step-by-step explanation:
given null hypothesis
H0=the individual has not taken steroids.
type 1 error-falsely rejecting the null hypothesis
⇒ actually the null hypothesis is true⇒the individual has not taken steroids.
but we rejected it ⇒our prediction is the individual has taken steroids.
typr II error- not rejecting null hypothesis when it has to be rejected
⇒actually null hypothesis is false ⇒the individual has taken steroids.
but we didnt reject⇒the individual has not taken steroids.
let us denote
the individual has taken steroids by 1
the individual has not taken steroids.by 0
predicted
1 0
actual 1 98.4% 1.6%
0 1.9% 98.1%
so for type 1 error
actual-0
predicted-1
therefore from above table we can see that probability of Type I error is 1.9%=0.019
so for type II error
actual-1
predicted-0
therefore from above table we can see that probability of Type I error is 1.6%=0.016
