MATHEMATICS COLLEGE

What appears to be the domain of the part of the exponential function graphed?A) 1 ≤ x ≤ 6
B) 1 ≤ ƒ(x) ≤ 6
C) –2 ≤ x ≤ 2
D) –2 ≤ ƒ(x) ≤ 2

Answers

Answer 1

Answer:

Answer:

Option (C)

Step-by-step explanation:

Domain of a function is represented by the 'x' values on the graph.

Similarly, y-values of a function represents the range.

Therefore, options (B) and (D) will not be the answer as they are representing the function values or y-values.

Given segments starts form x = -2 and ends at x = 2 (Including these points).

Therefore, Domain of the segment will be [-2, 2] Or -2 ≤ x ≤2

Option (C) will be the correct option.

Answer 2

Answer:

Answer:

B) 1 ≤ ƒ(x) ≤ 6

Step-by-step explanation:

this is the right answer I got it right on usa test prep

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HELP!! Find the value of x in the triangle!! 20PTS!!

Answers

Answer:

x = 5°

Step-by-step explanation:

We know that in a triangle, the measure of an exterior angle is equal to the sum of its two remote interior angles, therefore:

7x + 4 + 61 = 20x

7x + 65 = 20x

13x = 65

x = 5°

Answer:

Solution given:

61°+(7x+4)°=20x [ exterior angle is equal to the sum of two opposite interior angle]

65+7x=20x

65=20x-7x

13x=65°

x= $(65)/(13)$ =5°

value of x=5°

Let X be a binomially distributed random variable with parameters n=500 and p=0.3. The probability that X is no larger than one standard deviation above its mean is closest to which of the following? a. 0.579
b. 0.869
c. 0.847
d. 0.680

Answers

Answer:

c. 0.847

Step-by-step explanation:

From the given information;

$X \sim Binomial (500,0.3)$

Given that n = 500 which is too large, binomial distribution can now be approximated to

$N ( \mu , \sigma^2)$

where;

$\mu = np$

$\mu =500 * 0.3$

$\mu =150$

$\sigma^2= np(1-p)$

$\sigma^2= 500 * 0.3(1-0.3)$

$\sigma^2= 150(0.7)$

$\sigma^2= 105\n$

∴

$P(X \leq \mu + \sigma ) = ( (X-\mu)/(\sigma) \leq 1)$

$P(X \leq \mu + \sigma ) = ( Z \leq 1)$

$P(X \leq \mu + \sigma ) =\Phi (1)$

From the z table

$P(X \leq \mu + \sigma ) =0.841$

Thus, our value is closest to the option c which 0.847

Billy wants to buy 657 mugs. The mugs are sold for a price of 3 for $4. How much must Billy pay for all the mugs?

Answers

$(657mugs)/(3) = 219$
219 * $4 = $876

875 is the answer to your question so have a nice day

Patti drives a car for 224 miles in 4 hours what is her constant speed

Answers

Answer:

Step-by-step explanation:

The answer is 56 mph

Answer: 56

224/4 = 56
56 x 4 = 224

3 days after the start of an experiment there were 484 bacteria in a culture. After 5 days there were 1135. Use a system of equations to determine the initial number of bacteria in the culture (c) and the k value for the growth

Answers

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.

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