Answer:
6.2 units
Step-by-step explanation:
The mnemonic SOH CAH TOA is intended to remind you of the relations between trig functions and sides of a right triangle.
<h3>Common segment</h3>
To find the value of x in this figure, we need to know the length of the common segment between the two triangles. That segment is opposite the 53° angle, so can be found using the sine relation:
Sin = Opposite/Hypotenuse
Opposite = Hypotenuse × Sin = 30·sin(53°) ≈ 23.9591
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<h3>Missing measure</h3>
The side marked x is adjacent to the marked angle in that triangle, so the relevant relation is ...
Cos = Adjacent/Hypotenuse
Adjacent = Hypotenuse × Cos = 23.9591·cos(75°) ≈ 6.20106
The length of the side marked x is about 6.2 units.
Answer:
The probability that the total weight of the passengers exceeds 4222 pounds is 0.0018
Step-by-step explanation:
The Central limit Theorem stays that for a large value of n (21 should be enough), the average distribution X has distribution approximately normal with mean equal to 182 and standard deviation equal to 30/√21 = 6.5465. Lets call W the standarization of X. W has distribution approximately N(0,1) and it is given by the formula
In order for the total weight to exceed 4222 pounds, the average distribution should exceed 4222/21 = 201.0476.
The cummulative distribution function of W will be denoted by 
. The values of can be found in the attached file.
Therefore, the probability that the total weight of the passengers exceeds 4222 pounds is 0.0018.
4 hundreds 13 tens 5 ones = 4*100+13*10+5*1=400+130+5= \boxed {535}
Answer:
-64
Step-by-step explanation:
-21--43=22
The answer is <span>C=6p3 + 29p2 + 22p – 21</span>
To calculate the product, we need to multiply each member of each multiplier:
(2p + 7)(3p2 + 4p – 3) = 2p · 3p² + 2p · 4p + 2p · -3 + 7 ·3p² + 7 · 4p + 7 · -3
= 6p³ + 8p² - 6p + 21p² + 28p - 21
= 6p³ + 8p² + 21p² + <span>28p - 6p -21
= </span>6p³ + 29p² + 22p - 21
Therefore, the product of (2p + 7)(3p2 + 4p – 3) is 6p³ + 29p² + 22p -21
