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Makovka662 [10]
3 years ago
15

Ok so this is so confusing The volume of water in a rectangular swimming pool can be modeled by the polynomial 2x3 _ 9x2 + 7x +

6. If the depth of the pool is given by the polynomial 2x + 1, what polynomials express the length and width of the pool? Why couldn't they just give me the volume?
Mathematics
1 answer:
Alika [10]3 years ago
6 0
The volume formula is V= l x L x H, l=width, L=Length, H= Depth, so 
2x3 _ 9x2 + 7x + 6 = l x L x (2x + 1), because H=(2x + 1), so 
l x L= (2x3 _ 9x2 + 7x + 6 )/ (2x + 1) = (2x3 _ 9x2 + 7x + 6 ) X [1/(2x + 1)]
case1: l= (2x3 _ 9x2 + 7x + 6 ) or L= 1/(2x + 1), case2: L= (2x3 _ 9x2 + 7x + 6 ) or l= 1/(2x + 1)
the why question:
perhaps there is similarity of value between volume and l, or volume and L
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Oil is often measured in barrels one barrel can hold the 42 gallons of oil how many gallons of oil can 7 barrels hold
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294 gallons of oil is the answer because you do 42 and mltiply it by 7
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3 years ago
Three brothers, Andy, Bob, and Curly, take a taxi together home from the
MrRa [10]

Answer:

See explanation

Step-by-step explanation:

not sure what class this is for but I\ll try. Each traveller has a 21 km drive so their fare should be equal.

method 1) each traveller can pay for 21 km worth when they get out

method 2) Curly can pay the full fare and Andy and Bob can pay him back later (they're brothers so this should be easy to manage)

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3 years ago
A random variable x follows a normal distribution with mean d and standard deviation o=2. It is known that x is less than 5 abou
Vaselesa [24]

Answer:

The mean of this distribution is approximately 3.96.

Step-by-step explanation:

Here's how to solve this problem using a normal distribution table.

Let z be the

\displaystyle z = \frac{x - \mu}{\sigma}.

In this question, x = 5 and \sigma = 2. The equation becomes

\displaystyle z = \frac{5 - \mu}{2}.

To solve for \mu, the mean of this distribution, the only thing that needs to be found is the value of z. Since

The problem stated that P(X \le 5) = 69.85\% = 0.6985. Hence, P(Z \le z) = 0.6985.

The problem is that the normal distribution tables list only the value of P(0 \le Z \le z) for z \ge 0. To estimate  z from P(Z \le z) = 0.6985, it would be necessary to find the appropriate

Since P(Z \le z) = 0.6985 and is greater than P(Z \le 0) = 0.50, z > 0. As a result, P(Z \le z) can be written as the sum of P(Z < 0) and P(0 \le Z \le z). Besides, P(Z < 0) = P(Z \le 0) = 0.50. As a result:

\begin{aligned}&P(Z \le z)\\ &= P(Z < 0) + P(0 \le Z \le z) \\ &= 0.50 + P(0 \le Z \le z)\end{aligned}.

Therefore:

\begin{aligned}&P(0 \le Z \le z) \\ &= P(Z \le z) - 0.50 \\&= 0.6985 - 0.50 \\&=0.1985 \end{aligned}.

Lookup 0.1985 on a normal distribution table. The corresponding z-score is 0.52. (In other words, P(0 \le Z \le 0.52) = 0.1985.)

Given that

  • z = 0.52,
  • x =5, and
  • \sigma = 2,

Solve the equation \displaystyle z = \frac{x - \mu}{\sigma} for the mean, \mu:

\displaystyle 0.52 = \frac{5 - \mu}{2}.

\mu = 5 - 2 \times 0.52 = 3.96.

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3 years ago
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