Answer:
Is
Step-by-step explanation:
Hello Friend,here is the solution for your question
<span>so the given function is </span>
y= √(-2cos²x+3cosx-1)
i.e = √[-2(cos²x-3/2+1/2)]
i.e = √[-2(cosx-3/4)²-9/16+1/2]
i.e. = √[-2(cos-3/4)²-1/16]
i.e. = √[1/8-3(cosx=3/4)²]-----------(1)
Now here in this equation is this quantity :-
<span>(cosx=3/4)²----------------(2) is to it's minimum value then the whole equation </span>
<span>i.e. = √[1/8-3(cosx=3/4)²] will be maximum and vice versa </span>
And we know that cosx-3/4 will be minimum if cosx=3/4
<span>therefore put this in (1) we get </span>
(cosx=3/4)²=0 [ cosx=3/4]
<span>hence the minimum value of the quantity (cosx=3/4)² is 0 </span>
<span>put this in equation (1) </span>
we get ,
i.e. = √[1/8-3(cosx=3/4)²]
=√[1/8-3(0)] [ because minimum value of of the quantity (cosx=3/4)² is 0 ]
=√1/8
=1/(2√2)
<span>this is the maximum value now to find the minimum value </span>
<span>since this is function of root so the value of y will always be ≥0 </span>
<span>hence the minimum value of the function y is 0 </span>
<span>Therefore, the range of function </span>y is [0,1/(2√2)]
__Well,I have explained explained each and every step,do tell me if you don't understand any step._
Answer:
Your answer is D.
Step-by-step explanation:
If you need explanation ask me in comment
Answer:
Kindly check explanation
Step-by-step explanation:
Given the following :
Group 1:
μ1 = 59.7
s1 = 2.8
n1 = sample size = 12
Group 2:
μ2 = 64.7
s2 = 8.3
n2 = sample size = 15
α = 0.1
Assume normal distribution and equ sample variance
A.)
Null and alternative hypothesis
Null : μ1 = μ2
Alternative : μ1 < μ2
B.)
USing the t test
Test statistic :
t = (m1 - m2) / S(√1/n1 + 1/n2)
S = √(((n1 - 1)s²1 + (n2 - 1)s²2) / (n1 + n2 - 2))
S = √(((12 - 1)2.8^2 + (15 - 1)8.3^2) / (12 + 15 - 2))
S = 6.4829005
t = (59.7 - 64.7) / 6.4829005(√1/12 + 1/15)
t = - 5 / 2.5108165
tstat = −1.991384
Decision rule :
If tstat < - tα, (n1+n2-2) ; reject the Null
tstat < t0.1,25
From t table :
-t0.1, 25 = - 1.3163
tstat = - 1.9913
-1.9913 < - 1.3163 ; Hence reject the Null
