Answer:
The answer is false
Step-by-step explanation:
In a sample above 30 obs like this the confidence interval is defined as
X+- t* (s/sqrt(n)) where X is the mean t the tvalue for a given confidence level, n the size of sample and s standar deviation.
To find de appropiate value of t we must see the T table where rows are degrees of freedom and columns significance level
The significance is obtained:
significance = 1 - confidence level = 1 - 0.9 = 0.10
Degrees of freedom (df) for the inteval are
df = n - 1 = 18 - 1 = 17
So we must look for the value of a t with 17 values and significance of 0.10 which in t table is 1.740 not 1.746 ( thats the t for 16 df)
Given:
1st term = 11
common difference = 6
f(x) = 11 + 6(x - 1)
f(18) = 11 + 6(18-1)
f(18) = 11 + 6(17)
f(18) = 11 + 102
f(18) = 113 number of seats in row 18.
Row
<span>
<span>
</span><span><span>
1 11 11
</span><span>2 11 6 17
</span>
<span>
3 17 6
23
</span>
<span>
4 23 6 29
</span>
<span>
5 29 6 35
</span>
<span>
6 35 6
41
</span>
<span>
7 41 6 47
</span>
<span>
8 47 6 53
</span>
<span>
9 53 6 59
</span>
<span>
10 59 6
65
</span>
<span>
11 65 6 71
</span>
<span>
12 71 6
77
</span>
<span>
13 77 6
83
</span>
<span>
14 83 6
89
</span>
<span>
15 89 6 95
</span>
<span>
16 95 6
101
</span>
<span>
17 101
6
107
</span>
<span>
18 107
6 113
</span></span></span>
Answer:
<u>The present age of the mother is 30 years old</u>
Step-by-step explanation:
Age of Jane = x
Age of Billy = 5x (Billy is five times as old as Jane)
Age of Sara = 15x (Their mother is now three times as old as Billy)
In 2 years Sara will be eight times as old as Jane, therefore:
15x + 2 = 8 (x + 2)
15x + 2 = 8x + 16
15x - 8x = 16 - 2
7x = 14
x = 14/7
x = 2 ⇒ 5x = 10x ⇒ 15x = 30
Jane's present age is 2, Billy's present age is 10 and Sara's present age is 30
<u>The present age of the mother is 30 years old</u>
The correct answer is B -8 just do the work :)
To answer this, we need the original equation.
