Answer:
Step-by-step explanation:
We want to determine a 90% confidence interval for the mean amount of time that teens spend online each week.
Number of sample, n = 41
Mean, u = 43.1 hours
Standard deviation, s = 5.91 hours
For a confidence level of 90%, the corresponding z value is 1.645. This is determined from the normal distribution table.
We will apply the formula
Confidence interval
= mean +/- z ×standard deviation/√n
It becomes
43.1 ± 1.645 × 5.91/√41
= 43.1 ± 1.645 × 0.923
= 43.1 ± 1.52
The lower end of the confidence interval is 43.1 - 1.52 =41.58
The upper end of the confidence interval is 43.1 + 1.52 =44.62
Therefore, with 90% confidence interval, the mean amount of time that teens spend online each week is between 41.58 and 44.62
Answer:
$68.80
Step-by-step explanation:
First you need to find 60% of 48 which is 28.8. Then you add the 28.8 and 48 to get a total of $68.80.
Hope this helped :)
════════ ∘◦❁◦∘ ════════
<h3>Answer : (-2,11)</h3>
════════════════════
<h3>Known that</h3>
y = -2x² - 8x + 3
════════════════════
<h3>Way to do + explanation</h3>
#First, find the x coordinate first by using the formula of -b/2a
x = -b/2a = -(-8)/2(-2) = 8/-4 = -2
#After getting the x, find the y
y = -2(-2)² - 8(-2) + 3
y = -2×4 +16 + 3
y = -8 + 16 + 3
y = 11
So the vertex is
(-2,11)
════════════════════
The answer is 3. -1/2 n + 1 1/2 n add the coefficients to get 1n or just n. then n-n is 0 what is left is 3.