For each <em>x</em> in the interval 0 ≤ <em>x</em> ≤ 5, the shell at that point has
• radius = 5 - <em>x</em>, which is the distance from <em>x</em> to <em>x</em> = 5
• height = <em>x</em> ² + 2
• thickness = d<em>x</em>
and hence contributes a volume of 2<em>π</em> (5 - <em>x</em>) (<em>x</em> ² + 2) d<em>x</em>.
Taking infinitely many of these shells and summing their volumes (i.e. integrating) gives the volume of the region:
Answer:
Yes, we reject the auto maker's claim.
Step-by-step explanation:
H0 : μ ≥ 20
H1 : μ < 20
Sample mean, xbar = 18 ;
Sample size, n = 36
Standard deviation, s = 5
At α = 0.01
The test statistic :
(xbar - μ) ÷ s /sqrt(n)
(18 - 20) ÷ 5/sqrt(36)
-2 /0.8333333
= - 2.4
Pvalue from test statistic : Pvalue = 0.00819
Pvalue < α
0.00819 < 0.01
Hence, we reject the Null
The answer is 2 1/2kg = <span>2500 g</span>
Answer:
D. 14.32
Step-by-step explanation:
The formula for standard deviation is:
The mean μ is 33.
The (x minus μ) squared are 361, 49, 49, 361 respectively.
So that the SD =
The present age of Jane is 45 years old and present age of her sister is 9 years old
<em><u>Solution:</u></em>
Let the present age of Jane be "x"
Let the present age of her sister be "y"
<em><u>Jane is 5 times older than her sister</u></em>
present age of Jane = 5(present age of her sister)
x = 5y ---------- eqn 1
<em><u>In 3 years, Jane’s sister will be 1/4 her age</u></em>
Age of sister after 3 years = 3 + y
Age of jane after 3 years = 3 + x
Age of sister after 3 years = 1/4(age of jane after 3 years)
Substitute eqn 1 in above equation
Substitute y = 9 in eqn 1
x = 5(9)
x = 45
Thus present age of Jane is 45 years old and present age of her sister is 9 years old