The probability will be found as follows:
z-score is given by:
z=(x-μ)/σ
where:
mean-μ=9.3
standard deviation-σ=1.1
but
σ/√n=1.1/√70=0.1315
thus the z-score will be:
z=(9.1-9.3)/0.1315
z=-1.5209
thus:
P(x<9.1)=P(z<-1.5209)=0.0643
Hi there! You have to remember these 6 basic Trigonometric Ratios which are:
- sine (sin) = opposite/hypotenuse
- cosine (cos) = adjacent/hypotenuse
- tangent (tan) = opposite/adjacent
- cosecant (cosec/csc) = hypotenuse/opposite
- secant (sec) = hypotenuse/adjacent
- cotangent (cot) = adjacent/opposite
- cosecant is the reciprocal of sine
- secant is the reciprocal of cosine
- cotangent is the reciprocal of tangent
Back to the question. Assuming that the question asks you to find the cosine, sine, cosecant and secant of angle theta.
What we have now are:
- Trigonometric Ratio
- Adjacent = 12
- Opposite = 10
Looks like we are missing the hypotenuse. Do you remember the Pythagorean Theorem? Recall it!
Define that c-term is the hypotenuse. a-term and b-term can be defined as adjacent or opposite
Since we know the value of adjacent and opposite, we can use the formula to find the hypotenuse.
- 10²+12² = c²
- 100+144 = c²
- 244 = c²
Thus, the hypotenuse is:

Now that we know all lengths of the triangle, we can find the ratio. Recall Trigonometric Ratio above! Therefore, the answers are:
- cosine (cosθ) = adjacent/hypotenuse = 12/(2√61) = 6/√61 = <u>(6√61) / 61</u>
- sine (sinθ) = opposite/hypotenuse = 10/(2√61) = 5/√61 = <u>(5√61) / 61</u>
- cosecant (cscθ) is reciprocal of sine (sinθ). Hence, cscθ = (2√61/10) = <u>√61/5</u>
- secant (secθ) is reciprocal of cosine (cosθ). Hence, secθ = (2√61)/12 = <u>√</u><u>61</u><u>/</u><u>6</u>
Questions can be asked through comment.
Furthermore, we can use Trigonometric Identity to find the hypotenuse instead of Pythagorean Theorem.
Hope this helps, and Happy Learning! :)
That would be the commutative property
Answer:
true
Step-by-step explanation:
because when we subtract 18 and 4, the answer is 12. Hence, the value is less than 18
Step-by-step explanation:
(x^4)^3=(x^3)^4 , true
=> x^(4×3) = x^(3×4) = x^12
13^4 x 13^7= (13^4)^7, false
13^(4+7) = 13^11
(13^4)^7 = 13^(4×7) = 13^28
y^5 x y^0/y^3=(y^2)^1 , true
y^5 x y^0/y^3 = y^(5+0-3) = y^2
(y^2)^1 = y^(2×1) = y^2
q^0 x q^5/q^2=(q^3)^2/q^3, true
q^0 x q^5/q^2= q^(0+5-2)= q^3
(q^3)^2/q^3 = q^(3×2-3) = q^3