Answer:
The term Accuracy means that how close our result to the original result.
Suppose we do any experiment in laboratory and we calculate mass = 7 kg but answer is mass = 15 kg then our answer is not accurate.
And the term Precision means how likely we get result like this.
Suppose we do any experiment in laboratory and we calculate mass five times and each time we get mass = 7 kg then our answer is precised but not accurate.
Even though the content of many alcohol blends doesn't affect engine driveability, using gasoline with alcohol in warm weather may cause: decrease in fuel economy.
Mark brainliest
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Answer:
Amount of gas still in cylinder = 28 pound
Explanation:
Given:
Amount of gas in cylinder = 50 pound
Amount of gas used in Ms. Jones system = 13 pound
Amount of gas used in client system = 9 pound
Find:
Amount of gas still in cylinder
Computation:
Amount of gas still in cylinder = Amount of gas in cylinder - Amount of gas used in Ms. Jones system - Amount of gas used in client system
Amount of gas still in cylinder = 50 - 13 - 9
Amount of gas still in cylinder = 28 pound
Answer:
0.264 ; 0.079
Explanation:
Given that:
Sample size, n = 100
Probability of being active, p = 1% = 1/100 = 0.01
Using the binomial probability relation :
P(x =x) = nCx * p^x * (1 - p)^(n - x)
Probability that more than 1 user will be active
P(x > 1) = 1 - [p(x=0) + p(x = 1)]
P(x = 0) = 100C0 * 0.01^0 * 0.99^100 = 0.366
P(x = 1) = 100C1 * 0.01^1 * 0.99^99 = 0.370
P(x > 1) = 1 - [0.366 + 0.370]
P(x > 1) = 0.264
2.)
Probability that more than 2 user will be active
P(x > 2) = 1 - [p(x=0) + p(x = 1) + p(x = 2)]
P(x = 0) = 100C0 * 0.01^0 * 0.99^100 = 0.366
P(x = 1) = 100C1 * 0.01^1 * 0.99^99 = 0.370
P(x = 2) = 100C2 * 0.01^2 * 0.99^98 = 0.185
P(x > 1) = 1 - [0.366 + 0.370 + 0.185]
P(x > 1) = 0.079
Answer:
a) Fb= 275.77 lb Fc= 142.75 lb
b) M = -779.97 lb.ft (i.e. 779.97 lb.ft in clockwise direction)
c) Fax = 195 lb
Fay = 337.75 lb
Fbx = 195 lb
Fby = 195 lb
Explanation:
Question: Three tugboats are used to turn a barge in a narrow channel. To avoid producing any net translation of the barge, the forces applied should be couples. The tugboat at point A applies a 390 lb force.
(a) Determine FB and FC so that only couples are applied.
(b) Using your answers to Part (a), determine the resultant couple moment that is produced.
(c) Resolve the forces at A and B into x and y components, and identify the pairs of forces that constitute couples.
Solution:
<u>For this problem Right hand side is positive X direction and Upwards is positive Y direction. Couples and moments will be considered positive in counterclockwise direction.</u>
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a) For no translation condition
∑ 
& ∑
Hence,
and
Inserting the value of 
and solving the remaining equations simultaneously yields (magnitudes),
b) Summing up moments
(i.e. 779.97 lb.ft clockwise)
c)
