Answer:
1 a.) 2t – 2
1 b.) –9z + 4
2 a.) –2v + 3
2 b.) –13k – 9
3 a.) 7w − 2
3 b.) 10s
4 a.) 5
4 b.) –3p + 11
5 a.) –29x + 9
b.) 15v – 60x – 2
a.) 25n + 2
b.) –15n
a.) –71x + 4
b.) –15b
Step-by-step explanation:
- <em>NOTE: I am kind of confused about the way this question is formatted, so I tried to answer everything I could decipher.</em>
1 a.) –2t + 5t – 2 – t → 2t – 2
1 b.) –5z + 4 – 4z → –9z + 4
2 a.) 8v + 3 – 10v → –2v + 3
2 b.) –6 – 6k – 3 – 7k → –13k – 9
3 a.) –6 – 3w + 4 + 10w → 7w − 2
3 b.) 9s + s → 10s
4 a.) 6 + 8b – 1 – 8b → 5
4 b.) 1p + 6 + 5 – 4p → –3p + 11
5 a.) 4x – 3x + 9 – 6x 5 → –29x + 9
b.) 9v + 6v – 2 – 10x 6 → 15v – 60x – 2
a.) n + 2 – (–4)n 6 → 25n + 2
b.) 6n – 3n 7 → –15n
a.) –4x – 4x + 4 – 9x 7 → –71x + 4
b.) –9b + (–6)b → –15b
About 492 feet per minute
9 • 54.68066 = 492.126
Rounding to the nearest whole number gets you 492
Answer:
c. 6.2 ± 2.626(0.21)
Step-by-step explanation:
We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.
The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So
df = 101 - 1 = 100
99% confidence interval
Now, we have to find a value of T, which is found looking at the t table, with 100 degrees of freedom(y-axis) and a confidence level of
. So we have T = 2.626
The confidence interval is:

In which
is the sample mean while M is the margin of error.
The distribution of the number of puppies born per litter was skewed left with a mean of 6.2 puppies born per litter.
This means that 
The margin of error is:

In which s is the standard deviation of the sample and n is the size of the sample.
Thus, the confidence interval is:

And the correct answer is given by option c.
Its 360 beacuse u multiply all of those numbers
Answer:
-12
Step-by-step explanation:
We start with the equation: 
Substitute the variables: 
Solve using order of operations (start with multiplication): 
And then addition/subtration: 