The answer is 12 m.
Because you're not studying!
Next time do your mom.
Answer:
0.1225
Step-by-step explanation:
Given
Number of Machines = 20
Defective Machines = 7
Required
Probability that two selected (with replacement) are defective.
The first step is to define an event that a machine will be defective.
Let M represent the selected machine sis defective.
P(M) = 7/20
Provided that the two selected machines are replaced;
The probability is calculated as thus
P(Both) = P(First Defect) * P(Second Defect)
From tge question, we understand that each selection is replaced before another selection is made.
This means that the probability of first selection and the probability of second selection are independent.
And as such;
P(First Defect) = P (Second Defect) = P(M) = 7/20
So;
P(Both) = P(First Defect) * P(Second Defect)
PBoth) = 7/20 * 7/20
P(Both) = 49/400
P(Both) = 0.1225
Hence, the probability that both choices will be defective machines is 0.1225
<h2>
Answer:</h2>
<h2>
Step-by-step explanation:</h2>
For a better understanding of this problem, see the figure below. Our goal is to find 
. Since:
and is a common side both for ΔMRN and ΔMQN, then by SAS postulate, these two triangles are congruent and:
By Pythagorean theorem, for triangle NQP:
Applying Pythagorean theorem again, but for triangle MQN:
Answer:
8.20in³
Step-by-step explanation:
Given V = πr²h
r is the radius = 1.5in
h is the height = 6in
thickness of wall of the cylinder dr = 0.04in
top and bottom thickness dh 0.07in+0.07in = 0.14in
To compute the volume, we will find the value of dV
dV = dV/dr • dr + dV/dh • dh
dV/dr = 2πrh
dV/dh = πr²
dV = 2πrh dr + πr² dh
Substituting the values into the formula
dV = 2π(1.5)(6)•(0.04) + π(1.5)²(6) • 0.14
dV = 2π (0.36)+π(1.89)
dV = 0.72π+1.89π
dV = 2.61π
dV = 2.61(3.14)
dV = 8.1954in³
Hence volume, in cubic inches, of metal in the walls and top and bottom of the can is 8.20in³ (to two dp)
