Answer:
Population mean = 7 ± 2.306 × 
Step-by-step explanation:
Given - A university researcher wants to estimate the mean number
of novels that seniors read during their time in college. An exit
survey was conducted with a random sample of 9 seniors. The
sample mean was 7 novels with standard deviation 2.29 novels.
To find - Assuming that all conditions for conducting inference have
been met, which of the following is a 94.645% confidence
interval for the population mean number of novels read by
all seniors?
Proof -
Given that,
Mean ,x⁻ = 7
Standard deviation, s = 2.29
Size, n = 9
Now,
Degrees of freedom = df
= n - 1
= 9 - 1
= 8
⇒Degrees of freedom = 8
Now,
At 94.645% confidence level
α = 1 - 94.645%
=1 - 0.94645
=0.05355 ≈ 0.05
⇒α = 0.5
Now,

= 0.025
Then,
= 2.306
∴ we get
Population mean = x⁻ ±
×
= 7 ± 2.306 × 
⇒Population mean = 7 ± 2.306 × 
-99610=1700-55g
-1700 -1700
-101310=-55g
---------- ------
-55 -55
1842=g
Answer:x=53/6 and y=-55/6
Step-by-step explanation:
Let the first number x
Second number b y
Sum of the numbers x+y=-1/3
x-y=18
Solve simultaneously
Using elimination method
We get
2y=-55/3
y=-55/6
Substitute for y in equation 1
x+55/6=18
x=18-55/6
x=53/6
Answer:
A. 9A² - 16B²
Step-by-step explanation:
(3A + 4B)(3A - 4B)
Distribute:
3A(3A - 4B)
4B(3A - 4B)
Multiply:
3A(3A - 4B)
3A(3A) + 3A(-4B)
9A² - 12AB
4B(3A - 4B)
4B(3A) + 4B(-4B)
12AB - 16B²
Combine like terms:
9A² - 12AB + 12AB - 16B² <== -12AB + 12AB cancels out
9A² - 16B²
Therefore, our final answer is 9A² - 16B²
Hope this helps!