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dedylja [7]
4 years ago
8

HELP ASAP (MATH)

Mathematics
1 answer:
alexdok [17]4 years ago
8 0

PART - A : tent - 1 volume = 400 cubic ft

Tent -2 volume = 384 cubic ft.

PART - B : I recommend Joe to make the tent -1 with base 10x10 ft. as it has more volume

<u>Step-by-step explanation:</u>

PART - A:

tent 1 :

 Base = 10 x 10ft

 Height= 12 ft

 Volume = (1/3) (10x 10) (12)

 = (1/3) (100) (12)

= 400 cubic ft.

Tent 2:

 Base = 12 x 12 ft

 Height= 8 ft.

 Volume = (1/3) (12 x 12) (8)

= 384 cubic ft.

PART - B:

I recommend Joe to make the tent -1 with base 10x10 ft. as it has more volume

 

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3 years ago
X+y=250 tickets<br> 9x+6y=2,100 dollars.<br> Solve for X and Y Using the Combination Method
Gala2k [10]

Answer:

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Step-by-step explanation:

3 0
3 years ago
PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!
juin [17]

Answer: (B) min = -14, max = -3

<u>Step-by-step explanation:</u>

The minimum is the y-value of the vertex of a positive (U-shaped) parabola. In the graph, the vertex is located at (4.5, -14). The y-value of the vertex is -14.

The maximum is the y-value of the vertex of a negative (∩-shaped) parabola. In the graph, the vertex is located at (1.5, -3). The y-value of the vertex is -3.

8 0
3 years ago
A random sample of 60 suspension helmets used by motorcycle riders and automobile race-car drivers was subjected to an impact te
pentagon [3]

Answer:

The 95% confidence interval on the true proportion of helmets of this type that would show damage from this test is (0.169, 0.397).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

For this problem, we have that:

n = 60, \pi = \frac{17}{60} = 0.283

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.283 - 1.96\sqrt{\frac{0.283*0.717}{60}} = 0.169

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.283 + 1.96\sqrt{\frac{0.283*0.717}{60}} = 0.397

The 95% confidence interval on the true proportion of helmets of this type that would show damage from this test is (0.169, 0.397).

6 0
3 years ago
How do I solve systems of equations? I was absent and the test is next week!
MrRa [10]
Well, there are three different ways to solve systems of equations:
a) Graphing
b) Substitution
c) Elimination

Graphing:
If your system of equations are written in slope-intercept form (y=mx+b "m" being the slope and "b" being the y-intercept) its quite easy.
The y-intercept is pretty much where your line touches the y-axis so if your first equation is: y=3x-2 then you will put your first point on (0,-2). Then from that point you follow the slope. if the slope is 3 then you'll rise 3 and run 1 (go three times up and one to the right and put your point right there) till you feel you have enough points to make a line. You do the same process with your second equation.

Substitution:
This is best when you have a system of equations in which an equation tells you what x or y are equal to. For example: if I have "4x+2=y" and "2y+3x= 26" you can "replace" your "2y" with the equation "4x+2=y" which gives you a brand new equation: " 2(4x+2)+3x=26". To solve this you do the distributive property and link like terms which gives you the equation "11x+4=26". Then you solve like a normal equation, isolating x and then you'll get the value of x (in this case x=2). Now you go to the first equation "4x+2=y" and replace "x" to its value (in this case it's 2) and you get y (in this example, y=10).

Elimination:
I think this one is one of the easiest methods. Let's say my systems of equations is "5x-6y=-32" & "3x+6y=48". If you take a look at y's coefficients, you see they are the same. So you can "cancel them out" like this:
      5x-6y=-32
   + 3x+6y=48
-----------------------
     8x=16  <-------- when you get this, you can solve like a normal equation
       x=2

Also, if you have, for example:
2x+6y=190 & 2x+3y=130
You can solve this by: canceling x or multiplying the second equation by 2 and canceling y.
I'm gonna show you how to do the second way (multiplying and canceling y)
My second equation is "2x+3y=130". To get an equivalent equation, I just have to multiply the whole equation by any number. If I want to cancel y, then I have to multiply by 2.
So "2x+3y=130" turns into "4x+6y=260"
Now you follow the same procedure as before:
            2x+6y=190
       -    4x+6y=260
-------------------------------
      -2x= -70
       2x=70--------> x=35
Then you pretty much follow the substitution method by replacing "x" to "35" in one of the equations.

        Hope this covers all you missed in class. If you have any questions you can comment them here or message me. :D Good luck in your exam!!!

5 0
3 years ago
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