1.) <span>the ratio of the note A to middle C = 440.0 / 261.6 = 1.6820 ≈ 1.6818
2.) </span><span>the ratio of the note D to middle C = 293.7 / 261.6 = 1.1227
3.) </span><span>D# = 293.6 x 1.0595 = 311.1
4.) </span><span>the ratio of the frequency of G to C is 1 : 262/392 = 1 : 0.6683 ≈ 3 : 2
5.) </span><span>the ratio of the frequency of E to C = 1 : 262/330 = 1 : 0.7939 ≈ 5 : 4
6.) The </span><span>component in a musical note is referred to as the pitch of a sound is the frequency.</span>
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>
<u></u>
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:
-x² + 8x - 7
Step-by-step explanation:
Step 1: Write expression
(7 - x)(x - 1)
Step 2: FOIL (First Outside Inside Last)
7x - 7 - x² + x
Step 3: Combine like terms
-x² + 8x - 7
