Answer:
A) (17 ; 550)
B) $17/item
C) 550
Step-by-step explanation:
First we must calculate the intersection point of the two lines. Since in that point <em>y</em> has the same value in both equations, we can obtain <em>x </em>by equalling the two equations and then using that value for obtaining <em>y</em>:
So the value of <em>x</em> in the intersection point is 17. We now use this value with either one of the equations to obtain <em>y</em><em>. </em>Let's use the supply equation:
So the intersection point is (17 ; 550)
Supply and demand are in equilibrium when the amount of items on supply are the same as the ones on demand. That is the point were the two lines intersect, which means the selling price is the <em>x</em> coordinate and the amount of items is the <em>y</em> coordinate, so that is a selling price of <em>$17/item</em> with a number of items of <em>550</em>.
Answer:
yea
Step-by-step explanation:
Answer:
the answer is linear function lol
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.
8/30
= (8/2) / (30/2)
= 4/15
8/30 in its simplest form is 4/15~
