The gradient of the original line is 4. For a perpendicular gradient, you use the negative reciprocal, which is -1/4. Using y - y1 = m(x - x1), you can solve that y - 5 = -(1/4)(x - 2).
Multiply through by -4, you get -4y + 20 = x - 2, which can be rearranged as x + 4y = 22
Answer:
Step-by-step explanation:
Let us assume that the length of fishes in the tank is normally distributed. Thus, x is the random variable representing the length of fishes in the tank. Since the population mean and population standard deviation are known, we would apply the formula,
z = (x - µ)/(σ/√n)
Where
x = sample mean
µ = population mean
σ = standard deviation
number of samples
From the information given,
µ = 15
σ = 4
n = 31
We want to find the probability that x is between (15 - 0.5) inches and (15 + 0.5) inches. The probability is expressed as
P(14.5 ≤ x ≤ 15.5)
For x = 14.5
z = (14.5 - 15)/(4/√31) = - 0.7
Looking at the normal distribution table, the probability corresponding to the z score is 0.2420
For x = 15.5
z = (15.5 - 15)/(4/√31) = 0.7
Looking at the normal distribution table, the probability corresponding to the z score is 0.7580
Therefore,
P(14.5 ≤ x ≤ 15.5) = 0.7580 - 0.2420 = 0.5160
Answer:
We use students' t distribution therefore degrees of freedom is v= n-2
Step-by-step explanation:
<u>Confidence Interval Estimate of Population Regression Co efficient β.</u>
To construct the confidence interval for β, the population regression co efficient , we use b, the sample estimate of β. The sampling distribution of b is normally distributed with mean β and a standard deviation σ.y.x / √(x-x`)². That is the variable z = b - β/σ.y.x / √(x-x`)² is a standard normal variable. But σ.y.x is not known so we use S.y.x and also student's t distribution rather than normal distribution.
t= b - β/S.y.x / √(x-x`)² = b - β/Sb [Sb = S.y.x / √(x-x`)²]
with v= n-2 degrees of freedom.
Consequently
P [ - t α/2< b - β/Sb < t α/2] = 1- α
or
P [ b- t α/2 Sb< β < b+ t α/2 Sb] = 1- α
Hence a 100( 1-α) percent confidence for β the population regression coefficient for a particular sample size n <30 is given by
b± t α/2 Sb
Using the same statistic a confidence interval for α can be constructed in the same way for β replacing a with b and Sa with Sb.
a± t α/2 Sa
Using the t statistic we may construct the confidence interval for U.y.x for the given value X0 in the same manner
Y~0 ± t α/2(n-2) SY~
Y~0= a+b X0
Step-by-step explanation:
answer for d is 3 over 8....
that's all I know from the qn and can help u... sorry
