What pet makes the loudest noise algebra 2
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The manager of a computer retails store is concerned that his suppliers have been giving him laptop computers with lower than average quality. His research shows that replacement times for the model laptop of concern are normally distributed with a mean of 4.2 years and a standard deviation of 0.6 years. He then randomly selects records on 38 laptops sold in the past and finds that the mean replacement time is 3.9 years. Assuming that the laptop replacement times have a mean of 4.2 years and a standard deviation of 0.6 years, find the probability that 38 randomly selected laptops will have a mean replacement time of 3.9 years or less.
Name/ Uid:1. In this problem, try to write the equations of the given surface in the specified coordinates.(a) Write an equation for the sphere of radius 5 centered at the origin incylindricalcoordinates.(b) Write an equation for a cylinder of radius 1 centered at the origin and running parallel to thez-axis inspherical coordinates.
2. The U.S public health service calculates that service health care costs have increased from $43 →> $46.5 -> $49.8 -> $53.63 per patient during the last four years, what has been the cost per patient over the four-year period?
What is the pattern in the values as the exponents increase?
rested. Mr. Peters started out at the surface. He descended 25 feet, rose 8 feet and
descended another 6 feet. Then he rested. Which person rested at a greater depth?
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Organized workspace
Check work/answer
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Answer:
Mr.Matthews
Step-by-step explanation:
Mr.Matthews: Started at -12
-12 - 8 + 7 - 13 = -20 + 7 - 13 = -13 - 13 = -26
Mr. Peters: Started at -25
-25 + 8 - 6 = -17 - 6 = -23
In this case, -26 is greater than -23.
Mr.Matthews rested at a greater depth.
Hope this helps :)
a) 841
b) 106
c) 3 051
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Final answer:
To illustrate numbers using base 10 blocks, you need to understand the place values and represent each digit using the appropriate base 10 block.
Explanation:
To illustrate numbers using base 10 blocks, we need to understand that each number is made up of different place values—ones, tens, hundreds, and so on. A base 10 block for the number 841 would consist of 8 hundreds blocks, 4 tens blocks, and 1 ones block. For the number 106, we would have 1 hundreds block, 0 tens blocks, and 6 ones blocks. Lastly, for the number 3,051, we would have 3 thousands blocks, 0 hundreds blocks, 5 tens blocks, and 1 ones block.
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pounds of seed to completely plant an
8
-acre field. How many acres can be planted per pound of seed?
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Answer is 20
Please mark me as Brainliest ......
B) which inequality is shown on the graph?
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B : the line goes though the numbers less than 10, and the closed point shows it can be equal to 10.
Hope this helps!
b) What is the probability that the class hangs Wisconsin's flag on Monday, Michigan's flag on Tuesday, and California's flag on Wednesday.?
c) What is the probability that Wisconsin's flag will be hung at least two of the three days?
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Answer:
a.) P(x = X) =
b.)
c.) 0.00118
Step-by-step explanation:
The sample space Ω = flags of all 50 states
a.) Any one of the flags is randomly chosen. Therefore we can write the
probability measure as P(x = X) = , for all the elements of the sample
space, that is for all x ∈ Ω.
b.) the probability that the class hangs Wisconsin's flag on Monday,
Michigan's flag on Tuesday, and California's flag on Wednesday
=
c.) the probability that Wisconsin's flag will be hung at least two of the three days
= Probability that Wisconsin's flag will be hung on two days + Probability that Wisconsin's flag will be hung on three days
= P(x = 2) + P(x = 3)
=
=
=
= 0.00118
Final answer:
The sample space for this experiment is all the possible combinations of flags from the 50 U.S. states for the three days. The probability of hanging Wisconsin's flag on Monday, Michigan's on Tuesday, and California's on Wednesday is 1/125,000. The probability of hanging Wisconsin's flag at least two of the three days is 294/125,000.
Explanation:
a) The sample space Ω for this experiment comprises of all possible combinations of flags from the 50 U.S. states for the three days. Hence, the total number of outcomes in the sample space Ω would be 50*50*50 = 125,000. Every outcome in this space is equally likely, so the probability measure P would assign a probability of 1/125,000 to each outcome.
b) As each day's choice is independent of the others and each state's flag is equally likely to be chosen, the probability that Wisconsin's flag is hung on Monday, Michigan's flag is hung on Tuesday, and California's flag is hung on Wednesday would be (1/50) * (1/50) * (1/50) = 1/125,000.
c) To find the probability that Wisconsin's flag will be hung at least two of the three days, we have to add the probabilities for the three situations where Wisconsin's flag is hung exactly twice plus the situation where Wisconsin's flag is hung all three days. The final probability would be [(3 * (1/50)² * (49/50)) + (1/50)³] = 294/125,000.
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