Answer:
5,340
Step-by-step explanation:
Hi there:)
Amount invested in stock
=5000×0.6=3000
Amount invested in a saving account
5000-3000=2000
The stock increases 9% in the first year
3,000×(1+0.09)=3,270
and loses 4% of its value the second year
3,270×(1−0.04)=3,139.2
Amount of a saving account after two years
2,000×(1+0.049)^(2)=2,200.8
the total amount gained during the 2 years
3,139.2+2,200.8=5,340...answer
Hope it helps
Compab=a.b/|a|
b=<0,1,−2√10>
a.b= 2√10
|a| = √10
a.b/|a|=2√10 / √10=2
Answer:
Step-by-step explanation:
hello :
100+(n-2)² = 149
100-100+(n-2)² = 149-100
(n-2)² = 49
(n-2)² - 49 =0 but 49=7²
(n-2)² - 7² =0 use identity : a²-b²=(a-b)(a+b)
(n-2-7)(n-2+7)=0
(n-9)(n+5)=0
n-9=0 or n+5=0
n=9 or n=-5
The function given is a quadratic function, so the graph will be a parabola. It'll look similar to the photo attached. The minimum cost will be at the vertex of the parabola because that is its lowest point! To find the x-value of the vertex (which is what the question is looking for), use the vertex formula: x = -b/2a. The variable b is the coefficient of the x term in the function, and the variable a is the coefficient of the x² term. In this case, a = 0.125 and b = -5.
x = -(-5)/2(0.125)
x = 5/0.25
x = 20
So, 20 gas grills should be produced each day to maintain minimum costs. Hope that helps! :)
Answer:
Cov(X, Y) =0.029.
Step-by-step explanation:
Given that :
The noise in a particular voltage signal has a constant mean of 0.9 V. that is μ = 0.9V ............(1)
Also, the two noise instances sampled τ seconds apart have a bivariate normal distribution with covariance.
0.04e–jτj/10 ............(2)
Having X and Y denoting the noise at times 3 s and 8 s, respectively, the difference of time = 8-3 = 5seconds.
That is, they are 5 seconds apart,
τ = 5 seconds..............(3)
Thus,
Cov(X, Y), for τ = 5seconds = 0.04e-5/10
= 0.04e-0.5 = 0.04/√e
= 0.04/1.6487
= 0.0292
Thus, Cov(X, Y) =0.029.
