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:
4 zeros
Step-by-step explanation:
The zeroes of
f
(
x
)
=
2
x
4
−
5
x
3
+
3
x
2
+
4
x
−
6
are
x
=
−
1
,
x
=
3
2
,
x
=
1
−
i
, and
x
=
1
+
i
.
You have to divide
its 9,165 divided by 13
= 205
Teddy it takes 205 trips
to transfer 9,165 in a plane that can hold 13 people per trip
<em>Hope I helped</em> ^-^
D=daniel
brother=b
d is 3 older than 2 times b
d=3+2b
d=17
subsitute
17=3+2b
subtract 3 from both sides
17-3=3-3+2b
14=0+2b
14=2b
didivde both side by 2
7=b
daniels age in terms of his brother age is
d=3+2b