Answer:
(53.812 ; 58.188) ; 156
Step-by-step explanation:
Given that :
Sample size (n) = 51
Mean (m) = 56
Standard deviation (σ) = 9.5
α = 90%
Using the relation :
Confidence interval = mean ± Error
Error = Zcritical * (standard deviation / sqrt (n))
Zcritical at 90% = 1.645
Error = 1.645 * (9.5 / sqrt(51))
Error = 1.645 * 1.3302660
Error = 2.1882877
Hence,
Confidence interval :
Lower boundary = 56 - 2.1882877 = 53.8117123
Upper boundary = 56 + 2.1882877 = 58.1882877
Confidence interval = (53.812 ; 58.188)
2.)
Margin of Error (ME) = 1.25
α = 90%
Sample size = ((Zcritical * σ) / ME)^2
Zcritical at 90% = 1.645
Sample size = ((1.645 * 9.5) / 1.25)^2
Sample size = (15.6275 / 1.25)^2
Sample size = 12.502^2 = 156.3000
Sample size = 156
Answer:
Step-by-step explanation:
r = 21/42 = 1/2 ( also 10.5 / 21 = 1/2)
This is a Geometric Sequence with first term 42 and common ratio 1/2.
Answer: Option D. 16
Solution:
If LM is a midsegment of IJK, it is joining the midpoint of the sides IJ and IK, and it's half the length of the base of the triangle (JK), then:
L is the midpoint of the side IJ, and divides it into two congruent parts:
IL=LJ
Replacing IL by 7x and LJ by 3x+4:
7x=3x+4
Solving for x: Subtracting 3x both sides of the equation:
7x-3x=3x+4-3x
Subtracting:
4x=4
Dividing both sides of the equation by 4:
4x/4=4/4
Dividing:
x=1
Then we can determine the length of LM:
LM=2x+6
Replacing x by 1 in the equation above:
LM=2(1)+6
LM=2+6
LM=8
and because LM is half the length of the base of the triangle (JK)
LM=(1/2) JK
Replacing LM by 8:
8=(1/2) JK
Multiplying both sides of the equation by 2:
2(8)=2(1/2) JK
16=(2/2) JK
16=(1) JK
16=JK
JK=16
Step-by-step explanation:
The answer to your question is (D) 16. This is the answer because if count by 16's it goes 16, 24, 36, etc. since 24 is a multiple its could also be x
Answer:
The total number of cups in arranged in an hexagonal area = 19 cups
Step-by-step explanation:
The pack the most circles within an area, the arrangement with the densest packing is the hexagonal lattice structure similar to the bee's honeycomb as has been proved Gauss and Fejes Toth.
Therefore, we pack the circles in an hexagonal lattice structure in an assumed hexagonal area where we have;
The longest straight line is five cups the next on either side are four cups and the final line on either side has three cups
The total number of cups = 3 + 4 + 5 + 4 + 3 = 19 cups
The total number of cups = 19 cups.
