Given data, cos(A - B) = cosAcosB + sinAsinB
<span>let, A=60' and B=30' ( here the ' sign bears degree) </span>
<span>L.H.S = cos(A - B) </span>
<span>=cos (60'-30) ( using value of A and B ) </span>
<span>=cos30' </span>
<span>= sqrt3/2 </span>
<span>R.H.S= cosAcosB + sinAsinB </span>
<span>=cos60' cos30' + sin60' sin30' </span>
<span>= 1/2* sqrt3/2+ sqrt3/2*1/2 </span>
<span>= sqrt3/4 + sqrt3/4 </span>
<span>=2 sqrt3/4 </span>
<span>= sqrt3/2 </span>
<span>so L.H.S =R.H.S or cos(A - B) = cosAcosB + sinAsinB</span>
Answer:
B. 1.2×10^-3 mm
Step-by-step explanation:
What constitutes an "appropriate" unit depends on the use of the number. Since we're not told of any particular use, we might choose the smallest of the available units: mm.
1.2×10^-3 mm
_____
<em>Additional comment</em>
The most appropriate unit would be the one that eliminates the power-of-ten multiplier:
= 1.2 µm
Answer:
V = a³/8
Step-by-step explanation:
The volume of the original cube is the cube of the side length:
V = a³
When the side length is reduced to half its former value, the new volume is ...
V = (a/2)³ = a³/2³
V = a³/8 . . . . volume of the new cube
7 x 10^1 = 70
7 x 10^2 = 700
7 x 10^3 = 7000
Answer:
Null Hypothesis: H_0: \mu_A =\mu _B or \mu_A -\mu _B=0
Alternate Hypothesis: H_1: \mu_A >\mu _B or \mu_A -\mu _B>0
Here to test Fertilizer A height is greater than Fertilizer B
Two Sample T Test:
t=\frac{X_1-X_2}{\sqrt{S_p^2(1/n_1+1/n_2)}}
Where S_p^2=\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2}
S_p^2=\frac{(14)0.25^2+(12)0.2^2}{15+13-2}= 0.0521154
t=\frac{12.92-12.63}{\sqrt{0.0521154(1/15+1/13)}}= 3.3524
P value for Test Statistic of P(3.3524,26) = 0.0012
df = n1+n2-2 = 26
Critical value of P : t_{0.025,26}=2.05553
We can conclude that Test statistic is significant. Sufficient evidence to prove that we can Reject Null hypothesis and can say Fertilizer A is greater than Fertilizer B.
