Answer:
(115.2642, 222.7358).
Step-by-step explanation:
Given data:
type A: n_1=60, xbar_1=1827, s_1=168
type B: n_2=180, xbar_2=1658, s_2=225
n_1 = sample size 1, n_2= sample size 2
xbar_1, xbar_2 are mean life of sample 1 and 2 respectively. Similarly, s_1 and s_2 are standard deviation of 1,2.
a=0.05, |Z(0.025)|=1.96 (from the standard normal table)
So 95% CI is
(xbar_1 -xbar_2) ± Z×√[s1^2/n1 + s2^2/n2]
=(1827-1658) ± 1.96×sqrt(168^2/60 + 225^2/180)
= (115.2642, 222.7358).
Answer:
Part a) Daniel's age is 2 years
Part b) Kevin's age is 8 years
Step-by-step explanation:
<u><em>The question is </em></u>
Part a) How old is Daniel?
Part b) How old is Kevin?
Let
x ----> Kevin's age
y ----> Daniel's age
we know that
-----> equation A
----> equation B
Equate equation A and equation B
solve for y
therefore
Daniel's age is 2 years
<em>Find the value of x</em>
substitute the value of y in any of the two equations
therefore
Kevin's age is 8 years
Answer : 
(1) 
Use FOIL method to mulitply (3x – 5)(3x – 5)
(2) 
Use FOIL method to mulitply (3x - 5)(3x + 5)
(3) 
Use FOIL method to mulitply -(3x + 5)(3x + 5)
(4) 
Use FOIL method to mulitply -(3x + 5)(3x - 5)
So equation 2 is true
PART1:
First Combination:
Pizza ($7) + Chicken Strips ($6) + Biscuits ($3) + Grapes ($4) = $20
Second Combination:
Dog Food ($13) + Bread ($3) + Crackers ($2) + Broccoli ($2) = $20
Third Combination:
Shampoo ($4) + Tissues ($3) + Pizza ($7) + Eggs ($3) + Biscuits ($3) = $20
PART 2:
First Combination:
$7.20 + $5.70 + $2.90 + $3.70 = $19.60
No, I wouldn’t have gone over the limit
Second Combination:
$13.40 + $3.50 + $2.00 + $1.90 = $20.80
Yes, I would have gone over the limit
Third Combination:
$3.50 + $2.60 + $7.20 + $2.50 + $2.90 = $18.70
No, I wouldn’t have gone over the limit
Hope this helps!!
The answer is -9x + 6 2/3.
