The answer is 4 square root 5 or -4 square root 5
Can someone please help with this math question?
1 answer:
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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
Answer:
19 in.³
Step-by-step explanation:
We have been given that Jay created a model of a building from two rectangular prism. One building has a volume of 15 Cubic inches. The model had a total volume of 34 in.³ We are asked to find the volume of the other building.
We know that total volume of model will be equal to sum of volume of two buildings.
Therefore, the volume of the other building is 19 in.³