Answer:
(15,595, 16,805)
Step-by-step explanation:
We have to:
m = 16.2, sd = 3.75, n = 150
m is the mean, sd is the standard deviation and n is the sample size.
the degree of freedom would be:
n - 1 = 150 - 1 = 149
df = 149
at 95% confidence level the t is:
alpha = 1 - 95% = 1 - 0.95 = 0.05
alpha / 2 = 0.05 / 2 = 0.025
now well for t alpha / 2 (0.025) and df (149) = t = 1,976
the margin of error = E = t * sd / (n ^ (1/2))
replacing:
E = 1,976 * 3.75 / (150 ^ (1/2))
E = 0.605
The 95% confidence interval estimate of the popilation mean is:
m - E <u <m + E
16.2 - 0.605 <u <16.2 + 0.605
15,595 <u <16,805
(15,595, 16,805)
Answer:
9
Step-by-step explanation:
394,717.330 rounds to 400,000
All estimating problems make the assumption you are familar with your math facts, addition and multiplication. Since students normally memorize multiplication facts for single-digit numbers, any problem that can be simplified to single-digit numbers is easily worked.
2. You are asked to estimate 47.99 times 0.6. The problem statement suggests you do this by multiplying 50 times 0.6. That product is the same as 5 × 6, which is a math fact you have memorized. You know this because
.. 50 × 0.6 = (5 × 10) × (6 × 1/10)
.. = (5 × 6) × (10 ×1/10) . . . . . . . . . . . by the associative property of multiplication
.. = 30 × 1
.. = 30
3. You have not provided any clue as to the procedure reviewed in the lesson. Using a calculator,
.. 47.99 × 0.6 = 28.79 . . . . . . rounded to cents
4. You have to decide if knowing the price is near $30 is sufficient information, or whether you need to know it is precisely $28.79. In my opinion, knowing it is near $30 is good enough, unless I'm having to count pennies for any of several possible reasons.
