## Question 11:

### Part (a):

**Main Method:**

Line joining $(2, 3)$ to $(10, 200)$ has gradient:

$\frac{200-3}{10-2} = 24.625$ | M1 (AO 3.3) | Attempt to find the gradient of the line

$m = \frac{197}{8} = 24.625$, $c = -\frac{185}{4} = -46.25$ | A1, A1 (AO 1.1, 1.1) | Each value correct. Accept rounded to 2 or more s.f. Values may be implied by correct equation

Equation of the line is $N = 24.625t - 46.25$

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**Alternative Solution 1:**

Regression line (BC) for data set $(2, 3)$, $(10, 200)$ | M1 | Using the calculator to find the regression line (may be implied)

Regression line is $N = 24.625t - 46.25$

$m = 24.625$, $c = -46.25$ | A1, A1 | Each value correct. Accept rounded to 2 or more s.f. Values may be implied by correct equation

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**Alternative Solution 2:**

Simultaneous equations:

Using $(2, 3)$: $2m + c = 3$ and using $(10, 200)$: $10m + c = 200$ | M1 | Attempt to find and solve simultaneous equations (BC)

$m = \frac{197}{8} = 24.625$, $c = -\frac{185}{4} = -46.25$ | A1, A1 | Each value correct. Accept rounded to 2 or more s.f.

**[3]**

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### Part (b):

The number of calls increases by $24.625$ per day | B1 (AO 3.4) | FT their $m$. Condone missing units

**[1]**

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### Part (c):

When $t = 1$, $N = -21.625$ and $N$ cannot be negative | B1 (AO 3.5b) | Argument from a correct value for $N$. FT their equation

**[1]**

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### Part (d):

When $t = 2$, $N = e^{0.53 \times 2} = 2.886 \approx 3$ | M1 (AO 1.1a) | Substitution of both values into the model, soi

When $t = 10$, $N = e^{0.53 \times 10} = 200.33 \approx 200$

So this is a good model for these two points | E1 (AO 3.4) | Conclusion from correct values

**[2]**

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### Part (e):

Rate of increase $\frac{dN}{dt} = 0.53e^{0.53t}$ | M1 (AO 1.1a) | Using gradient of $e^{kx} = ke^{kx}$

When $t = 10$, $\frac{dN}{dt} = 0.53e^{0.53 \times 10}$ | M1 (AO 1.1a) | Substituting $t = 10$ into their expression

$106$ new calls per day | B1 (AO 3.4) | cao Allow for gradient found BC or numerical method

**[3]**