Answer:
The answer is 639.
Step-by-step explanation:
42.6 × 15
639
We have to divide 48 by 3 using distributive property.
Division using distributive property is used as it is much easier and accurate.
For dividing 48 by 3, we split first 48 as a multiple of 3 and 10
i.e. 48 = 30+18
Now divide again 18 as 3x5 +3
i.e. 48 = 30+15+3
By distributive property we have 48/3 = (30+15+3)/3 =30/3 +15/3 +3/3
Now easier to calclate this
30/3 = 10 : 15/3 =5 and 3/3 =1
Hence 48/3 = 10+5+1=16
Answer:
a and d
Step-by-step explanation:
The crate can carry any amount up to a pound. Since both b and c are over 1, they can immediately be eliminated. Since both a and d are lower than 1 they can be the answers. if the question tells you to pick all possible answers, pick both a and d but if it says pick one answer, pick a since it is the closest to one pound.
Answer:
The confidence interval for the population variance of the thicknesses of all aluminum sheets in this factory is Lower limit = 2.30, Upper limit = 4.83.
Step-by-step explanation:
The confidence interval for population variance is given as below:
![[(n - 1)\times S^{2} / X^{2} \alpha/2, n-1 ] < \alpha < [(n- 1)\times S^{2} / X^{2} 1- \alpha/2, n- 1 ]](https://tex.z-dn.net/?f=%5B%28n%20-%201%29%5Ctimes%20S%5E%7B2%7D%20%20%2F%20%20X%5E%7B2%7D%20%20%5Calpha%2F2%2C%20n-1%20%5D%20%3C%20%5Calpha%20%3C%20%5B%28n-%201%29%5Ctimes%20S%5E%7B2%7D%20%20%2F%20X%5E%7B2%7D%201-%20%5Calpha%2F2%2C%20n-%201%20%5D)
We are given
Confidence level = 98%
Sample size = n = 81
Degrees of freedom = n – 1 = 80
Sample Variance = S^2 = 3.23
![X^{2}_{[\alpha/2, n - 1]} = 112.3288\\\X^{2} _{1 -\alpha/2,n- 1} = 53.5401](https://tex.z-dn.net/?f=X%5E%7B2%7D_%7B%5B%5Calpha%2F2%2C%20n%20-%201%5D%7D%20%20%20%3D%20112.3288%5C%5C%5CX%5E%7B2%7D%20_%7B1%20-%5Calpha%2F2%2Cn-%201%7D%20%3D%2053.5401)
(By using chi-square table)
[(n – 1)*S^2 / X^2 α/2, n– 1 ] < σ^2 < [(n – 1)*S^2 / X^2 1 -α/2, n– 1 ]
[(81 – 1)* 3.23 / 112.3288] < σ^2 < [(81 – 1)* 3.23/ 53.5401]
2.3004 < σ^2 < 4.8263
Lower limit = 2.30
Upper limit = 4.83.
<h2><u>Part A:</u></h2>
Let's denote no of seats in first row with r1 , second row with r2.....and so on.
r1=5
Since next row will have 10 additional row each time when we move to next row,
So,
r2=5+10=15
r3=15+10=25
<u>Using the terms r1,r2 and r3 , we can find explicit formula</u>
r1=5=5+0=5+0×10=5+(1-1)×10
r2=15=5+10=5+(2-1)×10
r3=25=5+20=5+(3-1)×10
<u>So for nth row,</u>
rn=5+(n-1)×10
Since 5=r1 and 10=common difference (d)
rn=r1+(n-1)d
Since 'a' is a convention term for 1st term,
<h3>
<u>⇒</u><u>rn=a+(n-1)d</u></h3>
which is an explicit formula to find no of seats in any given row.
<h2><u>Part B:</u></h2>
Using above explicit formula, we can calculate no of seats in 7th row,
r7=5+(7-1)×10
r7=5+(7-1)×10 =5+6×10
r7=5+(7-1)×10 =5+6×10 =65
which is the no of seats in 7th row.
