The graph represents the cost y (in dollars) to open an online gaming account and buy x games. Write an equation in slope-interc
ept form that represents the graph
1 answer:
Answer:
y=10x+15
Step-by-step explanation:
The slope intercept form is represent by this equation
Where m is the slope, and b is the y intercept( when x is zero).
Looking at the graph, we already found the y intercept.
When x=0, y is at 15 so our y intercept is 15 so our equation updated look like
Now let find the slope
Slope is
Let use two arbitrary points to find the slope
Let use (3,45) and 6,75).
So our slope is 10 so our equation look like
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Problem 1, part A
X = number of times Roberto throws the baseball in the strike zone
X can take on the following values: 0, 1, 2, 3
p = 0.72 = probability of getting the ball in the strike zone (for any given independent trial)
n = 3 = sample size = number of times the baseball is thrown
Binomial Probabilities:
P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(X = 0) = (3 C 0)*(0.72)^(0)*(1-0.72)^(3-0)
P(X = 0) = (3 C 0)*(0.72)^(0)*(0.28)^(3)
P(X = 0) = (1)*(0.72)^(0)*(0.28)^3
P(X = 0) = (1)*(1)*(0.021952)
P(X = 0) = 0.021952
P(X = 0) = 0.022 <--- this value is added to the table (next to k = 0)
P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(X = 1) = (3 C 1)*(0.72)^(1)*(1-0.72)^(3-1)
P(X = 1) = (3 C 1)*(0.72)^(1)*(0.28)^(2)
P(X = 1) = (3)*(0.72)^(1)*(0.28)^2
P(X = 1) = (3)*(0.72)*(0.0784)
P(X = 1) = 0.169344
P(X = 1) = 0.169 <--- this value is added to the table (next to k = 1)
P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(X = 2) = (3 C 2)*(0.72)^(2)*(1-0.72)^(3-2)
P(X = 2) = (3 C 2)*(0.72)^(2)*(0.28)^(1)
P(X = 2) = (3)*(0.72)^(2)*(0.28)^1
P(X = 2) = (3)*(0.5184)*(0.28)
P(X = 2) = 0.435456
P(X = 2) = 0.435 <--- this value is added to the table (next to k = 2)
P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(X = 3) = (3 C 3)*(0.72)^(3)*(1-0.72)^(3-3)
P(X = 3) = (3 C 3)*(0.72)^(3)*(0.28)^(0)
P(X = 3) = (1)*(0.72)^(3)*(0.28)^0
P(X = 3) = (1)*(0.373248)*(1)
P(X = 3) = 0.373248
P(X = 3) = 0.373 <--- this value is added to the table (next to k = 3)
The table will look like what you see in the attached image
-----------------------------
Problem 1, part B
Refer to the table made in part A above. Add up the values in the second column, the P(X = k) column, that correspond to k values of 1 or larger. So basically everything but the first item which corresponds to k = 0
0.169+0.435+0.373 = 0.977
So the probability of at least one of the baseballs hits the strike zone is 0.977
=========================================
Problem 2
Convert each raw x score to a z score
Company A
z = (x-mu)/sigma
z = (260-276)/5.8
z = -2.759
----------------
Company B
z = (x-mu)/sigma
z = (260-252)/3.4
z = 2.353
----------------
The z scores are -2.759 and 2.353 for company A and company B respectively. The value -2.759 is further away from zero compared to 2.353, so company A has a lower chance of producing 260 nails. This is because company B has x = 260 closer to the mean (than compared to company A
X = number of times Roberto throws the baseball in the strike zone
X can take on the following values: 0, 1, 2, 3
p = 0.72 = probability of getting the ball in the strike zone (for any given independent trial)
n = 3 = sample size = number of times the baseball is thrown
Binomial Probabilities:
P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(X = 0) = (3 C 0)*(0.72)^(0)*(1-0.72)^(3-0)
P(X = 0) = (3 C 0)*(0.72)^(0)*(0.28)^(3)
P(X = 0) = (1)*(0.72)^(0)*(0.28)^3
P(X = 0) = (1)*(1)*(0.021952)
P(X = 0) = 0.021952
P(X = 0) = 0.022 <--- this value is added to the table (next to k = 0)
P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(X = 1) = (3 C 1)*(0.72)^(1)*(1-0.72)^(3-1)
P(X = 1) = (3 C 1)*(0.72)^(1)*(0.28)^(2)
P(X = 1) = (3)*(0.72)^(1)*(0.28)^2
P(X = 1) = (3)*(0.72)*(0.0784)
P(X = 1) = 0.169344
P(X = 1) = 0.169 <--- this value is added to the table (next to k = 1)
P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(X = 2) = (3 C 2)*(0.72)^(2)*(1-0.72)^(3-2)
P(X = 2) = (3 C 2)*(0.72)^(2)*(0.28)^(1)
P(X = 2) = (3)*(0.72)^(2)*(0.28)^1
P(X = 2) = (3)*(0.5184)*(0.28)
P(X = 2) = 0.435456
P(X = 2) = 0.435 <--- this value is added to the table (next to k = 2)
P(X = k) = (n C k)*(p)^(k)*(1-p)^(n-k)
P(X = 3) = (3 C 3)*(0.72)^(3)*(1-0.72)^(3-3)
P(X = 3) = (3 C 3)*(0.72)^(3)*(0.28)^(0)
P(X = 3) = (1)*(0.72)^(3)*(0.28)^0
P(X = 3) = (1)*(0.373248)*(1)
P(X = 3) = 0.373248
P(X = 3) = 0.373 <--- this value is added to the table (next to k = 3)
The table will look like what you see in the attached image
-----------------------------
Problem 1, part B
Refer to the table made in part A above. Add up the values in the second column, the P(X = k) column, that correspond to k values of 1 or larger. So basically everything but the first item which corresponds to k = 0
0.169+0.435+0.373 = 0.977
So the probability of at least one of the baseballs hits the strike zone is 0.977
=========================================
Problem 2
Convert each raw x score to a z score
Company A
z = (x-mu)/sigma
z = (260-276)/5.8
z = -2.759
----------------
Company B
z = (x-mu)/sigma
z = (260-252)/3.4
z = 2.353
----------------
The z scores are -2.759 and 2.353 for company A and company B respectively. The value -2.759 is further away from zero compared to 2.353, so company A has a lower chance of producing 260 nails. This is because company B has x = 260 closer to the mean (than compared to company A