Plug in 1 for n and solve, then plug in 2, and so on until you get to 3(6)-1=17
Answer:
The data is skewed to the bottom and contains an outlier.
Step-by-step explanation:
1. Test for outlier
An outlier is a point that is more than 1.5IQR below Q1 or above Q3.
IQR = Q3 - Q1 = 74 - 51 = 23
1.5 IQR = 1.5 × 23 = 34.5
51 - 15 = 36 > 1.5IQR
The point at 15 is an outlier.
2. Test for normal distribution
The median is not in the middle of the box.
Rather, it cuts the box into two unequal parts, so the data does not have a normal distribution.
3. Test for skewness
The longer part is to the left of the median, so the data is skewed left.
Answer:
24 per day
Step-by-step explanation:
'x' = days, so since 24 is with x, the rate of change is 24
Answer:
See in attachment and Mark as brainleist please.
Answer:
a) W₁ = 78400 [J]
b)Wt = 82320 [J]
Step-by-step explanation:
a) W = ∫ f*dl general expression for work
If we have a chain with density of 10 Kg/m, distributed weight would be
9.8 m/s² * 10 kg = mg
Total length of th chain is 40 m, and the function of y at any time is
f(y) = (40 - y ) mg where ( 40 - y ) is te length of chain to be winded
At the beggining we have to wind 40 meters y = 0 at the end of the proccess y = 40 and there is nothing to wind then:
f(y) = mg* (40 - y )
W₁ = ∫f(y) * dy ⇒ W₁ = ∫₀⁴⁰ mg* (40 - y ) dy ⇒ W₁ = mg [ ∫₀⁴⁰ 40dy - ∫₀⁴⁰ ydy
W₁ = mg [ 40*y |₀⁴⁰ - 1/2 * y² |₀⁴⁰ ⇒ W₁ = mg* [ 40*40 - 1/2 (40)² ]
W₁ = mg * [1/2] W₁ = 10*9,8* ( 800 )
W₁ = 78400 [J]
b) Now we can calculate work to do if we have a 25 block and the chain is weightless
W₂ = ∫ mg* dy ⇒ W₂ = ∫₀⁴⁰ mg*dy ⇒ W₂ = mg y |₀⁴⁰
W₂ = mg* 40 = 10*9.8* 40
W₂ = 3920 [J]
Total work
Wt = W₁ + W₂ ⇒ Wt = 78400 + 3920
Wt = 82320 [J]
