Answer:
The fraction or percentage of the applicants that we would expect to have a score of 400 or above is 77.34%
Step-by-step explanation:
Scores are normally distributed with a mean of 460 and a standard deviation of 80. For a value x, the associated z-score is computed as 
, therefore, the z-score for 400 is given by 
. To compute the fraction of the applicants that we would expect to have a score of 400 or above, we should compute the probability P(Z > -0.75) = 0.7734, i.e., the fraction or percentage of the applicants that we would expect to have a score of 400 or above is 77.34%
Answer:
I believe it is jk the figure?
Answer:
y = 8x + 9
Step-by-step explanation:
Parallel lines have the same gradient,
So m = 8
y = 8x + c
When x = -1 ,y = 1
1 = 8(-1) + c
c = 9
y = 8x + 9
Answer:
y-30 = 3(x-1)
Step-by-step explanation:
Point-slope form of an equation is
y-y1 = m(x-x1) where (x1,y1 ) is a point and m is the slope
We know on day 1 they exercised 30 minutes (1,30) where x is the number of days and y is the minutes. The rate increased 3 minutes each day so m = 3
y-30 = 3(x-1)
We have proven that the trigonometric identity [(tan θ)/(1 - cot θ)] + [(cot θ)/(1 - tan θ)] equals 1 + (secθ * cosec θ)
<h3>How to solve Trigonometric Identities?</h3>
We want to prove the trigonometric identity;
[(tan θ)/(1 - cot θ)] + [(cot θ)/(1 - tan θ)] = 1 + sec θ
The left hand side can be expressed as;
[(tan θ)/(1 - (1/tan θ)] + [(1/tan θ)/(1 - tan θ)]
⇒ [tan²θ/(tanθ - 1)] - [1/(tan θ(tanθ - 1)]
Taking the LCM and multiplying gives;
(tan³θ - 1)/(tanθ(tanθ - 1))
This can also be expressed as;
(tan³θ - 1³)/(tanθ(tanθ - 1))
By expansion of algebra this gives;
[(tanθ - 1)(tan²θ + tanθ.1 + 1²)]/[tanθ(tanθ(tanθ - 1))]
Solving Further gives;
(sec²θ + tanθ)/tanθ
⇒ sec²θ * cotθ + 1
⇒ (1/cos²θ * cos θ/sin θ) + 1
⇒ (1/cos θ * 1/sin θ) + 1
⇒ 1 + (secθ * cosec θ)
Read more about Trigonometric Identities at; brainly.com/question/7331447
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