Answer:
The answer is false
Step-by-step explanation:
In a sample above 30 obs like this the confidence interval is defined as
X+- t* (s/sqrt(n)) where X is the mean t the tvalue for a given confidence level, n the size of sample and s standar deviation.
To find de appropiate value of t we must see the T table where rows are degrees of freedom and columns significance level
The significance is obtained:
significance = 1 - confidence level = 1 - 0.9 = 0.10
Degrees of freedom (df) for the inteval are
df = n - 1 = 18 - 1 = 17
So we must look for the value of a t with 17 values and significance of 0.10 which in t table is 1.740 not 1.746 ( thats the t for 16 df)
Trinomial 2x² + 4x + 4.
It's of the form ax²+bx+c and it's discriminant is Δ=b² - 4.a.c
(in our case Δ = 4² - (4)(2)(4) → Δ = - 32
We know that: x' = -1 + i and x" = -1 - i
If Δ > 0 we have 2 rational solutions x' and x"
If Δ = 0 we have1 rational solution x' = x"
If Δ < 0 we have 2 complex solutions x' and x", that are conjugate
In our example we have Δ = - 16 then <0 so we have 2 complex solutions
That are x'= [-b+√Δ]/2.a and x" = [-b-√Δ]/2.a
x' =
Its B I think maybe if it’s wrong am sorry
The main formula is A=P(1+r/n)^nt, where A=amount of $ in the account at the specified time, P=principal (amount originally invested), r=interest rate, expressed as a decimal number, t=time, in years, of the investment, and n=number of times the account is compounded annually.
In our equation:
P=$11,600
r=7.25%=.0725
t=17 years
n=1 (compounded annually)
A= 11600(1+.[0725/1])^(1*17)
=11600(1+.0725)^17
=11600(1.0725)^17
=11600(3.286654969)
A=$38125.20
Answer:
$1.85
Step-by-step explanation:
$7.75 is $1.85 greater then $5.90
