Answer:
- multiplying a multi-digit number by itself several times (finding the power of a number)
- finding a square root
- statistical calculations
Step-by-step explanation:
We don't know what your introduction tells you, but the above-listed operations are ones I choose to use a calculator for. I also use a calculator for ordinary arithmetic, such as division by numbers with 2 digits or more. (It is simply faster and requires no scratch paper.)
If statistical calculations are not done with a calculator, they at least require the availability of suitable tables.
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All of these operations can be done by hand without a calculator, and were in times passed. Lifetimes of effort were involved in generating some of the original math tables for statistics, trig, logarithms, and other functions readily evaluated using a modern calculator.
Answer:
You would have 6 ducks
Step-by-step explanation:
Answer:
a) <u>0.4647</u>
b) <u>24.6 secs</u>
Step-by-step explanation:
Let T be interval between two successive barges
t(t) = λe^λt where t > 0
The mean of the exponential
E(T) = 1/λ
E(T) = 8
1/λ = 8
λ = 1/8
∴ t(t) = 1/8×e^-t/8 [ t > 0]
Now the probability we need
p[T<5] = ₀∫⁵ t(t) dt
=₀∫⁵ 1/8×e^-t/8 dt
= 1/8 ₀∫⁵ e^-t/8 dt
= 1/8 [ (e^-t/8) / -1/8 ]₀⁵
= - [ e^-t/8]₀⁵
= - [ e^-5/8 - 1 ]
= 1 - e^-5/8 = <u>0.4647</u>
Therefore the probability that the time interval between two successive barges is less than 5 minutes is <u>0.4647</u>
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b)
Now we find t such that;
p[T>t] = 0.95
so
t_∫¹⁰ t(x) dx = 0.95
t_∫¹⁰ 1/8×e^-x/8 = 0.95
1/8 t_∫¹⁰ e^-x/8 dx = 0.95
1/8 [( e^-x/8 ) / - 1/8 ]¹⁰_t = 0.95
- [ e^-x/8]¹⁰_t = 0.96
- [ 0 - e^-t/8 ] = 0.95
e^-t/8 = 0.95
take log of both sides
log (e^-t/8) = log (0.95)
-t/8 = In(0.95)
-t/8 = -0.0513
t = 8 × 0.0513
t = 0.4104 (min)
so we convert to seconds
t = 0.4104 × 60
t = <u>24.6 secs</u>
Therefore the time interval t such that we can be 95% sure that the time interval between two successive barges will be greater than t is <u>24.6 secs</u>
Answer:
x= 0 , 
, 
Step-by-step explanation:
Given, equation is ![\sqrt[3]{15x-1}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B15x-1%7D)
+ ![\sqrt[3]{13x+1}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B13x%2B1%7D)
= ![4\sqrt[3]{x}](https://tex.z-dn.net/?f=4%5Csqrt%5B3%5D%7Bx%7D)
. →→→ (1)
Now, by cubing the equation on both sides, we get
( + )³ = ()³
⇒ (15x-1) + (13x+1) + 3×× ( + ) = 64 x.
⇒ 28x + 3×× () = 64x.
(since from (1), + = )
⇒ 12× × (![\sqrt[3]{x}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%7D)
)= 36x.
⇒ 3x = ![\sqrt[3]{(15x-1)(13+1)(x)}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B%2815x-1%29%2813%2B1%29%28x%29%7D)
.
Now, once again cubing on both sides, we get
(3x)³ = ()³.
⇒ 27x³ = (15x-1)(13x+1)(x).
⇒ 27x³ = 195x³ + 2x² - x
⇒ 168x³ + 2x² - x = 0
⇒ x(168x² + 2x -1) = 0
⇒ by, solving the equation we get ,
x = 0 ; x = ; x =
therefore, solution is x= 0 , ,
Answer:
The perpendicular bisector meets any line at an angle which is not 90°
Step-by-step explanation:
