Part a.
The domain is the set of x values such that 
, basically x can be equal to -1/2 or it can be larger than -1/2. To get this answer, you solve 
for x (subtract 1 from both sides; then divide both sides by 2). I set 2x+1 larger or equal to 0 because we want to avoid the stuff under the square root to be negative.
If you want the domain in interval notation, then it would be 
which means the interval starts at -1/2 (including -1/2) and then it stops at infinity. So technically it never stops and goes on forever to the right.
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Part b.
I'm going to use "sqrt" as shorthand for "square root"
f(x) = sqrt(2x+1)
f(10) = sqrt(2*10+1) ... every x replaced by 10
f(10) = sqrt(20+1)
f(10) = sqrt(21)
f(10) = 4.58257569 which is approximate
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Part c.
f(x) = sqrt(2x+1)
f(x) = sqrt(2(x)+1)
f(x+2a) = sqrt(2(x+2a)+1) ... every x replaced by (x+2a)
f(x+2a) = sqrt(2x+4a+1) .... distribute
we can't simplify any further
-2 " -1,5 : -3-2 " -4/3 " -1 " 0 " 1 " 3/2 " 2
hv ju huy
Answer: (a) There is a clear outlier in the data.
Step-by-step explanation:
The statement relates that sample= 24
Confidence interval= 80.2 to 89.8.
Now the data does not mention anywhere the standard deviation which is an important parameter in using the t-procedure rather we assume that standard deviation is not known. So, the most worrying part is the skewness and the presence of strong outliers in the t-procedure.
So, the option a) is correct meaning a clear outlier in the data.
Answer:
0.4077
Step-by-step explanation:
It's important to use parenthesis to avoid confusion.
9^(2x) = 6
Use exponent properties:
(9^2)^x = 6
81^x = 6
Take log of both sides
ln(81^x) = ln(6)
Use log properties:
x ln(81) = ln(6)
Solve for x:
x = ln(6) / ln(81)
x ≈ 0.4077
Answer:
total money=rupees 50
3/5 of 50= rupees 30
money he spent = rupees 30
money he's left with =50-30=20
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