Answer:
sadd
Step-by-step explanation:
B
<span>The line that slopes downward (in blue) is the boundary line y = -x+5. The inequality y >= -x+5 will have a solid boundary line with shading above this line.
The red line sloping upward is the boundary line y = 3x-4. The inequality y <= 3x-4 means we shade below the red line
The two shaded regions from the inequalities will overlap to get region C.
Answer: Region C</span>
D. is the answer to your question, welcome :3
Answer:
Option 3x+5y=29 and -3x-12y=-48 is the system of equations equivalent to the given system of equations 3x+5y=29 and x+4y=16
Step-by-step explanation:
Given system of equations is 3x+5y=29 and x+4y=16
To find the equivalent system of equations to the given system of equations :
Option 3x+5y=29 and -3x-12y=-48 is the system of equations represents the given system of equations.
Because we can write the given equations as below
3x+5y=29 and x+4y=16
x+4y=16 rewritting as below
When multiply the above equation into (-3) we get
as same as the equation x+4y=16 so we can say that they are equivalent
Therefore Option 3x+5y=29 and -3x-12y=-48 is the system of equations represents the given system of equations
Therefore 3x+5y=29 and -3x-12y=-48 is the system of equations equivalent to the given system of equations 3x+5y=29 and x+4y=16
Answer:
Step-by-step explanation:
Since the incubation times are approximately normally distributed, we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = incubation times of fertilized eggs in days
µ = mean incubation time
σ = standard deviation
From the information given,
µ = 19 days
σ = 1 day
a) For the 20th percentile for incubation times, it means that 20% of the incubation times are below or even equal to 19 days(on the left side). We would determine the z score corresponding to 20%(20/100 = 0.2)
Looking at the normal distribution table, the z score corresponding to the probability value is - 0.84
Therefore,
- 0.84 = (x - 19)/1
x = - 0.84 + 19 = 18.16
b) for the incubation times that make up the middle 97% of fertilized eggs, the probability is 97% that the incubation times lie below and above 19 days. Thus, we would determine 2 z values. From the normal distribution table, the two z values corresponding to 0.97 are
1.89 and - 1.89
For z = 1.89,
1.89 = (x - 19)/1
x = 1.89 + 19 = 20.89 days
For z = - 1.89,
- 1.89 = (x - 19)/1
x = - 1.89 + 19 = 17.11 days
the incubation times that make up the middle 97% of fertilized eggs are
17.11 days and 20.89 days
