Answer:
- m∠A ≈ 53.13°
- m∠B ≈ 73.74°
- m∠C ≈ 53.13°
Step-by-step explanation:
An altitude to AC bisects it and creates two congruent right triangles. This lets you find ∠A = ∠C = arccos(6/10) ≈ 53.13°.
Since the sum of angles of a triangle is 180°, ∠B is the supplement of twice this angle, so is about 73.74°.
m∠A = m∠C ≈ 53.13°
m∠B ≈ 73.74°
_____
The mnemonic SOH CAH TOA reminds you of the relation between the adjacent side, hypotenuse, and trig function of an angle:
Cos = Adjacent/Hypotenuse
If the altitude from B bisects AC at X, triangle AXB is a right triangle with side AX adjacent to the angle A, and side AB as the hypotenuse. AX is half of AC, so has length 12/2 = 6, telling you the cosine of angle A is AX/AB = 6/10.
A diagram does not have to be sophisticated to be useful.
Answer:
see explanation
Step-by-step explanation:
Multiply both sides by 9 to eliminate the fraction
9K = 5(F + 459.67) ← divide both sides by 5
F + 459.67 = 
K ( subtract 459.67 from both sides )
F = K - 459.67
Add up the exponents of i : 0+1+2+3+4 = 10. So we have i^10.
This separates as follows: (i^4)^2 *i^2, which is equal to (1)^2 * (-1) = -1.
Answer:
The answer is c) 761.0
Step-by-step explanation:
Mathematical hope (also known as hope, expected value, population means or simply means) expresses the average value of a random phenomenon and is denoted as E (x). Hope is the sum of the product of the probability of each event by the value of that event. It is then defined as shown in the image, Where x is the value of the event, P the probability of its occurrence, "i" the period in which said event occurs and N the total number of periods or observations.
The variance of a random variable provides an idea of the dispersion of the random variable with respect to its hope. It is then defined as shown in the image.
Then you first calculate E [x] and E [
], and then be able to calculate the variance.
![E[x]=0*\frac{1}{40} +10*\frac{1}{20} +50*\frac{1}{10} +100*\frac{33}{40}](https://tex.z-dn.net/?f=E%5Bx%5D%3D0%2A%5Cfrac%7B1%7D%7B40%7D%20%2B10%2A%5Cfrac%7B1%7D%7B20%7D%20%2B50%2A%5Cfrac%7B1%7D%7B10%7D%20%2B100%2A%5Cfrac%7B33%7D%7B40%7D)
![E[x]=0+\frac{1}{2} +5+\frac{165}{2}](https://tex.z-dn.net/?f=E%5Bx%5D%3D0%2B%5Cfrac%7B1%7D%7B2%7D%20%2B5%2B%5Cfrac%7B165%7D%7B2%7D)
E[X]=88
So <em>E[X]²=88²=7744</em>
On the other hand
![E[x^{2} ]=0^{2} *\frac{1}{40} +10^{2} *\frac{1}{20} +50^{2} *\frac{1}{10} +100^{2} *\frac{33}{40}](https://tex.z-dn.net/?f=E%5Bx%5E%7B2%7D%20%5D%3D0%5E%7B2%7D%20%2A%5Cfrac%7B1%7D%7B40%7D%20%2B10%5E%7B2%7D%20%2A%5Cfrac%7B1%7D%7B20%7D%20%2B50%5E%7B2%7D%20%2A%5Cfrac%7B1%7D%7B10%7D%20%2B100%5E%7B2%7D%20%2A%5Cfrac%7B33%7D%7B40%7D)
E[x²]=0+5+250+8250
<em>E[x²]=8505
</em>
Then the variance will be:
Var[x]=8505-7744
<u><em>Var[x]=761
</em></u>
Answer:
Yes, your answers are correct!
Step-by-step explanation:
