Answer:
-0.25 or 1/4
Step-by-step explanation:
Eh just know
The power(P) of the circuits is the product of the voltage (V) in volts and current(I) in amperes.
P = IV
Substituting the known values to the equation,
P = (250 amps) x (40 volts) = 10,000 Watts
Thus, the answer is 10,000 Watts.
<span>The solution for a system of equations is the value or values that are true for all equations in the system. The graphs of equations within a system can tell you how many solutions exist for that system. Look at the images below. Each shows two lines that make up a system of equations.</span>
<span><span>One SolutionNo SolutionsInfinite Solutions</span><span /><span><span>If the graphs of the equations intersect, then there is one solution that is true for both equations. </span>If the graphs of the equations do not intersect (for example, if they are parallel), then there are no solutions that are true for both equations.If the graphs of the equations are the same, then there are an infinite number of solutions that are true for both equations.</span></span>
When the lines intersect, the point of intersection is the only point that the two graphs have in common. So the coordinates of that point are the solution for the two variables used in the equations. When the lines are parallel, there are no solutions, and sometimes the two equations will graph as the same line, in which case we have an infinite number of solutions.
Some special terms are sometimes used to describe these kinds of systems.
<span>The following terms refer to how many solutions the system has.</span>
A≈8
V Volume
Using the formula
V=a3
Solving fora
a=V⅓=512⅓≈8
Therefore, to find the lenght of an edge of the cube, just find the cube root of the volume. In this case, the cube root of 512 is equal to 8.
Answer:
Given the mean = 205 cm and standard deviation as 7.8cm
a. To calculate the probability that an individual distance is greater than 218.4 cm, we subtract the probability of the distance given (i.e 218.4 cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) from 1. Therefore, we have 1- P(Z
). Using the Z distribution table we have 1-0.9573. Therefore P(X >218.4)= 0.0427.
b. To calculate the probability that mean of 15 (i.e n=15) randomly selected distances is greater than 202.8, we subtract the probability of the distance given (i.e 202.8cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) divided by the square root of mean (i.e n= 15) from 1. Therefore, we have 1- P(Z
). Using the Z distribution table we have 1-0.1378. Therefore P(X >202.8)= 0.8622.
c. This will also apply to a normally distributed data even if it is not up to the sample size of 30 since the sample distribution is not a skewed one.
Step-by-step explanation:
Given the mean = 205 cm and standard deviation as 7.8cm
a. To calculate the probability that an individual distance is greater than 218.4 cm, we subtract the probability of the distance given (i.e 218.4 cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) from 1. Therefore, we have 1- P(Z). Using the Z distribution table we have 1-0.9573. Therefore P(X >218.4)= 0.0427.
b. To calculate the probability that mean of 15 (i.e n=15) randomly selected distances is greater than 202.8, we subtract the probability of the distance given (i.e 202.8cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) divided by the square root of mean (i.e n= 15) from 1. Therefore, we have 1- P(Z). Using the Z distribution table we have 1-0.1378. Therefore P(X >202.8)= 0.8622.
c. This will also apply to a normally distributed data even if it is not up to the sample size of 30 since the sample distribution is not a skewed one.
