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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?

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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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.

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5 0

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, $\mu$ :

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

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

3 0

3 years ago