Answer:
CI = (98.11 , 98.49)
The value of 98.6°F suggests that this is significantly higher
Step-by-step explanation:
Data provided in the question:
sample size, n = 103
Mean temperature, μ = 98.3
°
Standard deviation, σ = 0.73
Degrees of freedom, df = n - 1 = 102
Now,
For Confidence level of 99%, and df = 102, the t-value = 2.62 [from the standard t table]
Therefore,
CI = 
Thus,
Lower limit of CI = 
or
Lower limit of CI = 
or
Lower limit of CI = 98.11
and,
Upper limit of CI = 
or
Upper limit of CI = 
or
Upper limit of CI = 98.49
Hence,
CI = (98.11 , 98.49)
The value of 98.6°F suggests that this is significantly higher and the mean temperature could very possibly be 98.6°F
Answer:
She'll have to pay 4.8USD to get the bike
Answer:
Step-by-step explanation:
Remember these rules for addition, substraction, multiplication or division:
Addition
+ + = +
- - = -
Big + number - = +
Big -number + = -
Substraction
change the sign!
And follow the rules for addition
Multiplication or division
+ + = +
- - = +
+ - = -
Now for the exercise
<u>- (
)</u>
<u />
Change the signal
<u>(
)</u>
Answer:
k = 36 and r = 8
Step-by-step explanation:
Given that r varies directly as p and inversely as q² then the equation relating them is
r = 
← k is the constant of variation
To find k use the condition r = 27 when p 3 and q = 2, thus
27 = 
( multiply both sides by 4 )
108 = 3k ( divide both sides by 3 )
k = 36
r = 
← equation of variation
When p = 2 and q = 3, then
r = 
= 
= 8
Answer:
See below.
Step-by-step explanation:
ax^2 + bx + c = 0
a(x^2 + b/a x) + c = 0
Completing the square:
a [ (x + b/2a)^2 - b^2/4a^2] + c = 0
a[ (x + b/2a )]^2 - b^2 / 4a + c = 0
a[ (x + b/2a )]^2 = b^2 / 4a - c
Dividing both sides by a:
(x + b/2a )^2 = b^2/4a^2 - c/a
Taking square roots of both sides:
x + b/2a = +/- √ (b^2/ 4a^2 - c/a)
x + b/2a = +/- √ [ ( b^2 - 4ac) / 4a^2 )]
x + b/2a = +/- √ ( b^2 - 4ac) / 2a
Subtracting b/2a from both sides and converting the right side to one fraction:
x = [- b +/- √ ( b^2 - 4ac] / 2a.
