Answer:
The time the patient expected to survive after diagnosis is 29 years.
Step-by-step explanation:
It is provided that the mean survival time after diagnosis for a certain disease is 15 years with a standard deviation of 5 years.
That is,

An individual's predicted survival time is <em>a</em> = 2.8 standard deviations beyond the mean.
Compute the time the patient expected to survive after diagnosis as follows:


Thus, the time the patient expected to survive after diagnosis is 29 years.
Answer:
6.4 Cajas
Step-by-step explanation:
320/50 = 6.4 Cajas
<h2>Answer </h2>
The length of UC is 18
<h2>Explanation </h2>
First we are going to find the length of JN; then we are subtracting from it the length of JU plus the length of CN.
We can infer from our picture that JN is 82 + 105, so JN = 187
We can also infer that JU = JH + HU
JU = 22 + 96
JU = 118
We can also infer that CN = 51
Now we can fin the length of UC:



We can conclude that the length of UC is 18.
Solution:
Given:

To get sin 240 degrees:
240 degrees falls in the third quadrant.
In the third quadrant, only tangent is positive. Hence, sin 240 will be negative.

Using the trigonometric identity;

Hence,

To get cos 240 degrees:
240 degrees falls in the third quadrant.
In the third quadrant, only tangent is positive. Hence, cos 240 will be negative.

Using the trigonometric identity;

Hence,

To get tan 240 degrees:
240 degrees falls in the third quadrant.
In the third quadrant, only tangent is positive. Hence, tan 240 will be positive.

Using the trigonometric identity;

Hence,

To get cosec 240 degrees:

To get sec 240 degrees:

To get cot 240 degrees:
Answer:
Below.
Step-by-step explanation:
We apply the quotient rule:
dy/dx = ( tanx * 3x^2 - x^3 * sec^2x) / (tan^2 x)
= x^2( 3tanx - xsec^2x) / tan^2x