Answer:
4.54 g/cm^3
Step-by-step explanation:
d = m/v
d = 454 g ÷ 100.0 cm^3
d = 4.54 g/cm^3
3 sig figs so this answer is complete
Answer:
Step-by-step explanation:
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Good evening ,
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2(7) + 5(1) + 3(-3) = 10
3(7) - (1) + 4(-3) = 8
5(7) - 2(1) + 7(-3) = 12
Then (7,1,-3) is a solution to system
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:)
<span>60
Sorry, but the value of 150 you entered is incorrect. So let's find the correct value.
The first thing to do is determine how large the Jefferson High School parking lot was originally. You could do that by adding up the area of 3 regions. They would be a 75x300 ft rectangle, a 75x165 ft rectangle, and a 75x75 ft square. But I'm lazy and another way to calculate that area is take the area of the (300+75)x(165+75) ft square (the sum of the old parking lot plus the area covered by the school) and subtract 300x165 (the area of the school). So
(300+75)x(165+75) - 300x165 = 375x240 - 300x165 = 90000 - 49500 = 40500
So the old parking lot covers 40500 square feet. Since we want to double the area, the area that we'll get from the expansion will also be 40500 square feet. So let's setup an equation for that:
(375+x)(240+x)-90000 = 40500
The values of 375, 240, and 90000 were gotten from the length and width of the old area covered and one of the intermediate results we calculated when we figured out the area of the old parking lot. Let's expand the equation:
(375+x)(240+x)-90000 = 40500
x^2 + 375x + 240x + 90000 - 90000 = 40500
x^2 + 615x = 40500
x^2 + 615x - 40500 = 0
Now we have a normal quadratic equation. Let's use the quadratic formula to find its roots. They are: -675 and 60. Obviously they didn't shrink the area by 675 feet in both dimensions, so we can toss that root out. And the value of 60 makes sense. So the old parking lot was expanded by 60 feet in both dimensions.</span>
Since you have y alone on one side of the equation......
Use substitution and plug in 1/2x-4 for y in the other equation.
Distribute
Combine like terms
Add 8 to both sides
Divide by 4
and 
Plug this value in for x
Simplify
Multiply everything by 8 and simplify
Divide.....
Your answers are:
Answer:
p(on schedule) ≈ 0.7755
Step-by-step explanation:
A suitable probability calculator can show you this answer.
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The z-values corresponding to the build time limits are ...
z = (37.5 -45)/6.75 ≈ -1.1111
z = (54 -45)/6.75 ≈ 1.3333
You can look these up in a suitable CDF table and find the difference between the values you find. That will be about ...
0.90879 -0.13326 = 0.77553
The probability assembly will stay on schedule is about 78%.
