Answer:
y≤ -1/4 - 7x/8
Step-by-step explanation:
Answer: Part a) 375 pounds, Part b) 8 words
Step-by-step explanation:
a)
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Since n represents the number of words there are, we can plug in 20 for n into the equation to get:
C = 15*20 + 75 pounds
C = 300 + 75 pounds
C = 375 pounds
b)
Since C represents the cost of the advert, we can plug in 195 for C into the equation to get:
195 = 15n + 75
Subtract 75 from both sides:
120 = 15n
Divide both sides by 15:
n = 8 words
Answer:
a. N=25
b. X[bar]= 60.52
c. Y[bar]= 106.72
d. SSx= 115.24
e. ∑X*∑Y = 4036684
f. SSxy= 202020.3296
g. √(SSx*SSy)= 449.46
Step-by-step explanation:
Hello!
Using the attached data you need to calculate some statistics.
a) N
The sample size is listed under the first column "subject" You can see that 25 subjects qhere studied so N=25.
b.
The mean of set X is equal to X[bar]= ∑X/n= 1513/25= 60.52
∑X is listed in the second table.
c.
The mean of ser Y is Y[bar]= ∑Y/n= 2668/25= 106.72
∑Y is listed in the second table.
d.
Sum of Squares of set X SSx= ∑X²-[(∑X)²/n]= 91682-[(1513)²/25]= 115.24
e.
∑X*∑Y =1513*2668= 4036684
f.
SSxy= (∑X²-[(∑X)²/n]) * (∑Y²-[(∑Y)²/n])= (91682-[(1513)²/25]) * (286482*[(2668)²/25])= 202020.3296
g.
√(SSx*SSy)= √(115.24*1753)= 449.46
I hope you have a SUPER day!
Answer:
Therefore the concentration of salt in the incoming brine is 1.73 g/L.
Step-by-step explanation:
Here the amount of incoming and outgoing of water are equal. Then the amount of water in the tank remain same = 10 liters.
Let the concentration of salt be a gram/L
Let the amount salt in the tank at any time t be Q(t).
Incoming rate = (a g/L)×(1 L/min)
=a g/min
The concentration of salt in the tank at any time t is = g/L
Outgoing rate =
Integrating both sides
[ where c arbitrary constant]
Initial condition when t= 20 , Q(t)= 15 gram
Therefore ,
.......(1)
In the starting time t=0 and Q(t)=0
Putting t=0 and Q(t)=0 in equation (1) we get
Therefore the concentration of salt in the incoming brine is 1.73 g/L
Answer:
(9x²)²
Step-by-step explanation:
Given the expression 81x⁴, to write the expression as a square of a monomial, first we will assign a variable to the expression.
y = 81x⁴
Then we take the square root of both sides of the expression
√y = √81x⁴
y^½ = √81 × √x⁴
y^½ = 9x²
Squaring both sides of the resulting equation to get y back
(y^½)² = (9x²)²
y = (9x²)²
The expression as a square of a monomial is (9x²)²