Answer: x>_3.2 OR x<_ -0.75
Step-by-step explanation: first break down your compound inequality. 5x-4>_12
You first cancel out your constants by adding 4 to both sides. Now you’re left with 5x>_16 then to cancel five you have to divide on both sides by five which equals to 3.2. Then, x>_ 3.2.
Next you do your second part, 12x+5<_-4
So first cancel out the constant of 5 by subtracting 5 on both sides, making the equation 12x<_-9. Now, you divide by 12 on both sides, making it -9/12. Which effectively is -0.75. Therefor, the answer being x<_ -0.75. Add the two together x>_3.2 OR x<_0.75
Answer:
Step-by-step explanation:
The value of x associated with maximum allowable length is:
The maximum packaging cost is:
The domain and range values for this function are, respectively:
Answer: E) Mean will decrease, standard deviation will decrease
Step-by-step explanation:
Initial mean = 81
Initial standard deviation = 9
New score = 78
Taking a look at the initial mean(averahe) score, which is 81 and the new score to be added to the initial scores, the initial average is greater than the new score. Hence, this will result in a decrease in the new mean score after adding the new score of 78.
Also, taking a look at the standard deviation which is 9, we can conclude that the variability in initial scores from the mean scores is high. However, the new score of 78 and the initial mean are very close, and tend to low variation. Hence, adding the new score will lead to a decrease in variability and hence a decrease in the value of standard deviation.
If you've started pre-calculus, then you know that the derivative of h(t)
is zero where h(t) is maximum.
The derivative is h'(t) = -32 t + 96 .
At the maximum ... h'(t) = 0
32 t = 96 sec
t = 3 sec .
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If you haven't had any calculus yet, then you don't know how to
take a derivative, and you don't know what it's good for anyway.
In that case, the question GIVES you the maximum height.
Just write it in place of h(t), then solve the quadratic equation
and find out what 't' must be at that height.
150 ft = -16 t² + 96 t + 6
Subtract 150ft from each side: -16t² + 96t - 144 = 0 .
Before you attack that, you can divide each side by -16,
making it a lot easier to handle:
t² - 6t + 9 = 0
I'm sure you can run with that equation now and solve it.
The solution is the time after launch when the object reaches 150 ft.
It's 3 seconds.
(Funny how the two widely different methods lead to the same answer.)
The answer is from AL2006
Answer:
A. 160 balls
Step-by-step explanation:
First we find the volume of each sphere.
diameter = d = 0.75 in.
radius = r = d/2 = 0.75/2 in. = 0.375 in.
V = (4/3)π(r^3) = (4/3)(3.14159)(0.375 in.)^3 = 0.221 in.^3
Because of the assumption about the space occupied by a sphere in random packing, we now calculate 190% of the volume of the sphere.
190%V = 1.9 * 0.221 in.^3 = 0.420 in.^3
Now we calculate the volume of the box.
V = LWH = 3 in. * 3 in. * 7.5 in. = 67.5 in.^3
To find out the number of balls that fit in the box, we divide the volume of the box by the randomly packed ball volume.
number of balls = (67.5 in.^3)/(0.420 in.^3) = 160.8
Answer: 160 balls
