How do I expand
r(r-r2) the 2 is meant to be r squared
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that correspond to 80 % degree of confidence and the sample size
n=15.
χ2L= χ L 2 =
χ2R= χ R 2 =
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Answer: this is very hard and i was wondering if there is more info
Step-by-step explanation:
Final answer:
The critical values for a Chi-square distribution with 80% confidence level and sample size of 15 (thus 14 degrees of freedom) are: χ2L = 18.475 and χ2R = 8.897.
Explanation:
The question is asking for the critical values for a Chi-square distribution which can be found using Chi-square tables usually found in statistics textbooks or online. The critical values refer to the boundary values for the acceptance region of a hypothesis test.
When trying to find the critical values for an 80% confidence level with 14 degrees of freedom (since degrees of freedom = n - 1 for sample, thus 15 - 1 = 14), you need the 90th percentile for the χ2L and the 10th percentile for the χ2R (since α = 20%, then α/2 = 10%).
Using a Chi-square table for these percentiles, we get:
χ2L = 18.475
χ2R = 8.897
Learn more about Chi-square Critical Values here:
#SPJ12
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Answer:
Similarities
1. Adding or subtracting a number from both sides.
2. Multiplying or dividing both sides by a positive number.
3. Applying an increasing function to both sides.
4. Simplifying one or both sides.
Differences
1. Multiplying or dividing both sides by a negative number changes the direction of an inequality. So we cannot multiply or divide both sides by a variable unless we know that the variable is either always positive or always negative. This is not a concern when solving equalities.
2. When applying a decreasing function to both sides of an inequality, the direction of the inequality changes.
3. Swapping both sides of an inequality also changes the direction of an inequality. Again we don't have to worry about this with equalities.
(-2, 0) and (-12, 0)
(4.0) and (6.0)
(12, 0) and (2.0)
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Answer:
(4.0) and (6.0)
Step-by-step explanation:
when the function cross the x- axis
the y-coordinate equals to zero
x2 - 10x + 24=0
x=6,4
y=0
Answer: C.) (4,0) and (6,0)
Step-by-step explanation:
The other person is correct
Edge. 2022
B. y = -4x + 2
C. y = 4x + 2
D. y = 2x - 2
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Answer:
Part 1: 359,007 ft³
Part 2: 216 times smaller
Part 3: 21600%
Step-by-step explanation:
Part 1:
The parameters for the tank are;
The radius of the tank = 70 feet
The volume of a sphere = 4/3·π·r³
Therefore, the volume of a quarter sphere = 1/4×The volume of a sphere
The volume of a quarter sphere = 1/4×4/3·π·r³ = π·r³/3
Plugging in the value for the radius gives
Volume = π×70³/3 = 114,333.33×3.14 = 359,006.7≈ 359,007 ft³.
Part 2:
The dimension of the scale model = 1/6 × Actual dimension
Therefore, we have the radius of the sphere of the scale model = 1/6 × 70
Which gives;
The radius of the sphere of the scale model = 35/3 = 11.67 feet
The volume of the scale model = π·r³/3 = (3.14×11.67³)/3 = 1662.07 ≈ 1662 ft³
The number of times smaller the scale model is than the actual volume = (Actual volume)/(Scale model) = (359,007 ft³)/(1662 ft³) = 216 times
The number of times smaller the scale model is than the actual volume = 216 times = (1/Scale of model)³ = (1/(1/6))³ = 6³.
Part 3:
The percentage of the mock-up, x, to the volume of the actual tank is given as follows
x/100 × 1662 = 359,007
∴ x = 216 × 100 = 21600%
The percentage of the mock-up, to the volume of the actual tank is 21600%.
Answer:
Part 1: 359,007 ft³
Part 2: 216 times smaller
Part 3: 21600%
Step-by-step explanation:
Part 1:
The parameters for the tank are;
The radius of the tank = 70 feet
The volume of a sphere = 4/3·π·r³
Therefore, the volume of a quarter sphere = 1/4×The volume of a sphere
The volume of a quarter sphere = 1/4×4/3·π·r³ = π·r³/3
Plugging in the value for the radius gives
Volume = π×70³/3 = 114,333.33×3.14 = 359,006.7≈ 359,007 ft³.
Part 2:
The dimension of the scale model = 1/6 × Actual dimension
Therefore, we have the radius of the sphere of the scale model = 1/6 × 70
Which gives;
The radius of the sphere of the scale model = 35/3 = 11.67 feet
The volume of the scale model = π·r³/3 = (3.14×11.67³)/3 = 1662.07 ≈ 1662 ft³
The number of times smaller the scale model is than the actual volume = (Actual volume)/(Scale model) = (359,007 ft³)/(1662 ft³) = 216 times
The number of times smaller the scale model is than the actual volume = 216 times = (1/Scale of model)³ = (1/(1/6))³ = 6³.
Part 3:
The percentage of the mock-up, x, to the volume of the actual tank is given as follows
x/100 × 1662 = 359,007
∴ x = 216 × 100 = 21600%
The percentage of the mock-up, to the volume of the actual tank is 21600%.