s +h = 1070 ( total number of downloads)
Rewrite as s =1070 - h
2.9s + 4.4h = 4393 ( total download size)
Substitute:
2.9(1070-h) + 4.4h = 4393
Simplify:
3103 - 2.9h + 4.4h = 4393
Combine like terms:
3103 + 1.5h = 4393
Subtract 3103 from each side:
1.5h = 1290
Divide both sides by 1.5:
h = 1290 / 1.5
h = 860
Replace h with 860 in first equation to solve for s:
s + 860 = 1070
s = 1070 - 860
s = 210
There were 210 standard downloads and 860 high quality downloads
Answer:
<em>The weight of the parcel is approximately 6 kilograms.</em>
Step-by-step explanation:
The weight of the parcel is 12 lbs and 8 oz.
1 lb = 16 oz
So, 8 oz =
lb.
<u>That means, 12 lbs and 8 oz</u> = (12 + 0.5) lbs = 12.5 lbs
Now, 1 lb = 454 g
So, 12.5 lbs = (12.5 × 454) g = 5675 g
As, 1 kilogram = 1000 g
That means, 1 g = 0.001 kilogram
So, 5675 g
kilograms. <em>(Rounded to the nearest whole kilogram)</em>
Thus, the weight of the parcel is approximately 6 kilograms.
Answers:
bio = 258
sociology = 169
============================================================
Work Shown:
x = number of bio textbooks
y = number of sociology textbooks
x+y = 427 books sold total
y = x-89 since 89 less sociology textbooks were sold compared to bio
Apply substitution to get the following
x+y = 427
x+x-89 = 427 ... replace y with x-89
2x-89 = 427
2x = 427+89
2x = 516
x = 516/2
x = 258 bio textbooks were sold
y = x-89
y = 258-89
y = 169 sociology textbooks were sold
P(1) = P(-1)
P(1) = 3 - a + b
P(-1) = -3 + a + b
-> 3 - a + b = -3 + a + b
-> 3 - a + b + 3 - a - b = 0
-> 6 - 2a = 0
-> a = 3.
P(2) = 24 - 2a + b -> 24 - (2a - b) = 16 -> 2a - b = 8
-> 6 - b = 8
b = -2.
So, a = 3 and b = -2
Recheck : P(1) = 3 - 3 + (-2) = -2
P(-1) = -3 + 3 + (-2) = -2 => P(1) = P(-1) (true)
P(2) = 24 - 6 + (-2) = 16.
We know form our problem that the third day she biked 20 miles, so we have the point (3,20). We also know that <span>on the eighth day she biked 35 miles, so our second point is (8,35).
To relate our two point we are going to use the slope formula: </span>

We can infer form our points that

,

,

, and

. so lets replace those values in our slope formula:



Now that we have the slope, we can use the point-slope formula <span>determine the equation of the line that best fit the set for Maggie’s data.
Point-slope formula: </span>




We can conclude that the equation of the line that best fit the set for Maggie’s data is

.