²log(x+2)+²log(x-2)≤²log5
1 answer:
I believe you meant to write
?
If that's the case I'll solve the one I provided but I'll drop the base 2 to type it faster but you need to put it always!
Remember: log a + log b = log (a*b)
So log (x+2) + log (x-2) = log [(x+2)*(x-2)] = log (x^2 - 4)
Now back to the inequality:
log (x^2 - 4) <span>≤ log 5
Raise both sides as powers of 2 ( Since it's the base of your log)
Now,
x^2 - 4 </span><span>≤ 5
Add 4 both sides:
x^2 </span>≤ 9
Square root both sides
x ≤ +3 or x ≤ -3
Reject the -3 solution as it makes both (x + 2) and (x - 2) negative and a log can never have a negative value inside its brackets.
So x <span>≤ 3 But can never be less than 2 as well for the same previous reason.
Hope that helped.</span>
If that's the case I'll solve the one I provided but I'll drop the base 2 to type it faster but you need to put it always!
Remember: log a + log b = log (a*b)
So log (x+2) + log (x-2) = log [(x+2)*(x-2)] = log (x^2 - 4)
Now back to the inequality:
log (x^2 - 4) <span>≤ log 5
Raise both sides as powers of 2 ( Since it's the base of your log)
Now,
x^2 - 4 </span><span>≤ 5
Add 4 both sides:
x^2 </span>≤ 9
Square root both sides
x ≤ +3 or x ≤ -3
Reject the -3 solution as it makes both (x + 2) and (x - 2) negative and a log can never have a negative value inside its brackets.
So x <span>≤ 3 But can never be less than 2 as well for the same previous reason.
Hope that helped.</span>
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Answer:
a)
i. x=- - - - - -
ii. o=- - - - - -
iii. Sx=- - - - -
b) The random variable X signifies the weights of the bags of candies which are selected at random.
c) The statistics variable X is a measure on a sample that is used as an estimate of the population mean.X is the mean weight of the 16 bags of candies that were selected.
d) The distribution needed for this problem is normal distribution with parameters because the population standard deviation is known.
N(X, o/)
So, the distribution is
N(2, 0.12/)
e)
i. The 90% confidence interval for the population mean weight of the candies is 1.9589 , 2.0411
iii. The error bound is 0.0411
f)
i. The 98% confidence interval for the population mean weight of the candies is 1.9418 , 2.0582
iii. The error bound is 0.0582
g) The difference in the confidence intervals in part (f) and part (e) is of the level of confidence. The change is, the change in the area being calculated for the normal distribution. Therefore, the larger confidence level results in larger area and larger interval.
Hence the interval in part is larger than the one in part (e).
h) The interval in part (f) signifies that with 98% confidence that the population mean weight of the bag of candies lies between 1.942 ounce and 2.058 ounce.
Step-by-step explanation:
a)
i. x=- - - - - -
ii. o=- - - - - -
iii. Sx=- - - - -
b) The random variable X signifies the weights of the bags of candies which are selected at random.
c) The statistics variable X is a measure on a sample that is used as an estimate of the population mean.X is the mean weight of the 16 bags of candies that were selected.
d) The distribution needed for this problem is normal distribution with parameters because the population standard deviation is known.
N(X, o/)
So, the distribution is
N(2, 0.12/)
e)
i. The 90% confidence interval for the population mean weight of the candies is 1.9589 , 2.0411
iii. The error bound is 0.0411
f)
i. The 98% confidence interval for the population mean weight of the candies is 1.9418 , 2.0582
iii. The error bound is 0.0582
g) The difference in the confidence intervals in part (f) and part (e) is of the level of confidence. The change is, the change in the area being calculated for the normal distribution. Therefore, the larger confidence level results in larger area and larger interval.
Hence the interval in part is larger than the one in part (e).
h) The interval in part (f) signifies that with 98% confidence that the population mean weight of the bag of candies lies between 1.942 ounce and 2.058 ounce.
because