Read the information in the box below.Last year during the hurricane season, the amount of rain that fell on the Texas Gulf Coast in October was 14 inches, which was 312times more than the rainfall during September of the same year. How much rain fell during the month of September?

Rebecca tried to solve the problem but made a mistake in her calculations. You must analyze Rebecca's work, identify the error, and then correctly solve the problem. Then, create a model to justify your thinking.
Rebecca's Math Calculations
Step 1: 14×312
Step 2: 141×72
Step 3: 982=49
Solution: In September, 49 inches of rain fell.







Be sure to –
Identify the error that the student made
Answer the question prompt
Create a pictorial model that represents the problem situation and justifies your identification of the error and its solution
Explain how the model justifies the identified error and how to correct it

Step-by-step explanation:

k(a) = | -2a + 3 | - 1

k(3) = | -2(3) + 3 | - 1        replaced the a with 3

      = | -6 + 3 | - 1

      = | -3 | - 1

      =    3   - 1

      = 2

g(x) = 4²ˣ⁻¹ + 7

g(1) = 4²⁽¹⁾⁻¹ + 7                 replaced the x with 1

      = 4¹ + 7

      = 4 + 7

      = 11

f(x) = | 8x² - 5x + 3 |

f(-2) = | 8(-2)² - 5(-2) + 3 |        replaced the x with -2

      = | 8(4) - 10 + 3 |

      = | 12 - 10 + 3 |

      = | 5 |

      = 5

h(x) = -3x + 9

h(-1 + x) = -3(-1 + x) + 9          replaced the x with -1 + x

            = 3 - 3x + 9

            = 12 - 3x

f(n) = 5n - 1

f(-3n) = 5(-3n) - 1                 replaced the n with -3n

        = -15n - 1

Answer:

10 ft

Step-by-step explanation:

The word problem is basically asking the length of one corner of the rectangle to the other. We can split the rectangle up into two right triangles and we will use the Pythagorean theorem to find the hypotenuse or the length from one corner to another. A^2 + B^2 = C^2 A and B are the lengths of the rectangle, 8 and 6. 8^2 + 6^2 = C^2

8^2 = 64 and 6^2 = 36

64 + 36 = 100.

the square root of 100 is 10 so the length of the hose is `10.

Answer:

1479 g

Step-by-step explanation:

The volume of the bar is

V = l*w*h

  = 10*3*5

  = 150 cm^3

Now multiply by the density

150 cm^3 * 9.86 g/ cm^3 =1479 g

Answer:

• c = 135
• k = 0.42615

Step-by-step explanation:

We assume you want your model to be ...

  p = c·e^(kt)

Filling in (t, p) values of (3, 484) and (5, 1135), we have two equations in the two unknowns:

  484 = c·e^(3k)

  1135 = c·e^(5k)

Taking logs makes these linear equations:

  ln(484) = ln(c) +3k

  ln(1135) = ln(c) +5k

Subtracting the first equation from the second, we have ...

  ln(1135) -ln(484) = 2k

  k = ln(1135/484)/2 ≈ 0.42615

Using that value in the first equation, we find ...

  ln(484) = ln(c) +3(ln(1135/484)/2)

  ln(c) = ln(484) -(3/2)ln(1135/484)

  c = e^(ln(484) -(3/2)ln(1135/484)) ≈ 134.8

The initial number in the culture was 135, and the k-value is about 0.42615.

_____

I prefer to start with the model ...

  p = 484·(1135/484)^((t-3)/2)

Then the initial value is that obtained when t=0:

  c = 484·(1135/484)^(-3/2) = 134.778 ≈ 135

The value of k the log of the base for exponent t. It is ...

  ln((1135/484)^(1/2)) = 0.426152

This starting model matches the given numbers exactly. The transformation to c·e^(kt) requires approximations that make it difficult to match the given numbers.

__

For this model, the base of the exponent is the ratio of the two given population values. The exponent is horizontally offset by the number of days for the first count, and scaled by the number of days between counts. The multiplier of the exponential term is the first count. The model can be written directly from the given data, with no computation required.