Answer:
A = Heterozygous
B = Capital
C = Capital
G = Heterozygous
Step-by-step explanation:
Depends on which is dominant (Captital Letter)
If both are the same (lowercase or uppercase) eg TT, or tt They are capital
If different casings, they are Heterozygous.
(DUE TO Brainly disliking the word please refer Capital as H,o,m,o,z,y,g,o,u,s
Answer:
B
Step-by-step explanation:
y=-2x+2
y=mx+b
m is slope so.here m is -2
y=-2+10.passes through the point (3,-4)
because -4=-2*3+2
-4=-6+2
-4=-4
Answer:
Step-by-step explanation:
Answer:
![\bar X= \frac{20*20 +12*30 +7*10 +5*15 +6*10}{20+30+10+15+10}](https://tex.z-dn.net/?f=%20%5Cbar%20X%3D%20%5Cfrac%7B20%2A20%20%2B12%2A30%20%2B7%2A10%20%2B5%2A15%20%2B6%2A10%7D%7B20%2B30%2B10%2B15%2B10%7D)
![\bar X = \frac{965}{85}= 11.353](https://tex.z-dn.net/?f=%20%5Cbar%20X%20%3D%20%5Cfrac%7B965%7D%7B85%7D%3D%2011.353)
Step-by-step explanation:
For this case we have the following data:
Days to Maturity Dollar Value $ Millions)
20 20
12 30
7 10
5 15
6 10
For this case the definition of weigthed mean is given by:
![\bar X = \frac{\sum_{i=1}^n w_i x_i}{\sum_{i=1}^n w_i}](https://tex.z-dn.net/?f=%20%5Cbar%20X%20%3D%20%5Cfrac%7B%5Csum_%7Bi%3D1%7D%5En%20w_i%20x_i%7D%7B%5Csum_%7Bi%3D1%7D%5En%20w_i%7D)
Where
represent the weight of each value and
represent the observations (Days to maturity)
Our interest is the number of days so then our weighting variable would be the dollar Value, so then we can find the weighted mean like this:
![\bar X= \frac{20*20 +12*30 +7*10 +5*15 +6*10}{20+30+10+15+10}](https://tex.z-dn.net/?f=%20%5Cbar%20X%3D%20%5Cfrac%7B20%2A20%20%2B12%2A30%20%2B7%2A10%20%2B5%2A15%20%2B6%2A10%7D%7B20%2B30%2B10%2B15%2B10%7D)
![\bar X = \frac{965}{85}= 11.353](https://tex.z-dn.net/?f=%20%5Cbar%20X%20%3D%20%5Cfrac%7B965%7D%7B85%7D%3D%2011.353)
Answer:
labor fee: $15.15, quart of oil: $1.25
Step-by-step explanation:
You solve this by setting up two equations, one for each car.
First we have to define what variables we need and what they mean.
To be solvable, you need as many equations as you have variables.
Let's call the labor fee x and the cost of oil per quart y.
This means we'll be needing two equations to have any chance of solving this. Since we have one equation per car, this looks promising!
Car 1: x + 5y = 21.4
Car 2: x + 7y = 23.9
To solve these, you combine them into one equation by eliminating one variable:
Rewriting the first: x = 21.4 - 5y
So, we can substitute 21.4 - 5y for x in the second. Then we have an equation with only y:
21.4 - 5y + 7y = 23.9
2y = 23.9-21.4 = 2.5
y = 1.25
Plugging this value back into the rewritten first gives us the labour fee:
x = 21.4 - 5*1.25 = 15.15