Width = x
Length = x+18
Assuming the table is rectangular:
Area = x(x + 18)
Therefore:
x(x + 18) <span>≤ 175
x^2 + 18x </span><span>≤ 175
Using completing the square method:
x^2 + 18x + 81 </span><span>≤ 175 + 81
(x + 9)^2 </span><span>≤ 256
|x + 9| </span><span>≤ sqrt(256)
|x + 9| </span><span>≤ +-16
-16 </span>≤ x + 9 <span>≤ 16
</span>-16 - 9 ≤ x <span>≤ 16 - 9
</span>-25 ≤ x <span>≤ 7
</span><span>
But x > 0 (there are no negative measurements):
</span><span>
Therefore, the interval 0 < x </span><span>≤ 7 represents the possible widths.</span><span>
</span>
The answer is <span>The </span>root<span> of a number x is another number, which when multiplied by itself a given number of times, equals x. For example the second </span>root<span> of 9 is 3, because 3x3 = 9. The second </span>root<span> is usually called the square </span>root<span>. The third </span>root<span> is susually called the cube </span>root<span> See </span>Root<span> (of a number). because then/</span>
Let X be the number of female employee. Let n be the sample size, p be the probability that selected employee is female.
It is given that 45% employee are female it mean p=0.45
Sample size n=60
From given information X follows Binomial distribution with n=50 and p=0.45
For large value of n the Binomial distribution approximates to Normal distribution.
Let p be the proportion of female employee in the given sample.
Then distribution of proportion P is normal with parameters
mean =p and standard deviation = 
Here we have p=0.45
So mean = p = 0.45 and
standard deviation = 
standard deviation = 0.0642
Now probability that sample proportions of female lies between 0.40 and 0.55 is
P(0.40 < P < 0.45) = 
= P(-0.7788 < Z < 1.5576)
= P(Z < 1.5576) - P(Z < -0.7788)
= P(Z < 1.56) - P(Z < -0.78)
= 0.9406 - 0.2177
= 0.7229
The probability that the sample proportion of females is between 0.40 and 0.55 is 0.7229
<u>Answer-</u>
<u>Solution-</u>
Given expression is 
Applying Binomial Theorem
Here,
a = x, b = 2 and n = 4
So,
Expanding the summation
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>
