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2 years ago

1.

2 answers:

8 0

1. The answer is 76.

If the group all together is only being charged once per ride, and the whole group is going twice, then you will double the cost of each ride:

4.50 x 2 = 9
4 x 2 = 8
3.50 x 2 = 7

In total, the rides will cost $24. So, if you're paying with a $100 bill, you will receive $76 in change.

2.2 The answer is 58.

You will do the same procedure as the first question, but this time multiply the costs by 3.

5.50 x 3 = 16.5
2 x 3 = 6
6.50 x 3 = 19.5

The total will be 42. Subtracted from 100, you will get 58.

3.3 The answer is 2.

If you have 15 dollars and the coaster costs 5.50, the first time you ride it you will have 9.50 left. The second time, you will have 4 dollars left, and then you will not be able to pay for a third time.

4.4 The answer is 23.25

If you add up all the correlating costs of the tickets (6.75 + 4.50 + 4.50 + 10.50 + 10.50), you will get 36.75. Then you must subtract that from 60, which will give you 23.25 dollars in change.

5.5 The answer is 54.75

You will do the same procedure here as you did above. Add the corresponding costs together (you're done after this part, because the question isn't asking for how much change you'll receive).

6.6 The answer is 7.

If the tickets cost $10.50 and you have $80, then you want to divide 80 by 10.50. You will get 7.6. Since you can't buy a sixth of a ticket, your answer will be 7.

7.7 The answer is to pay separately.

In this question, you will add all the costs together (except the day pass) and compare it to the cost of the day pass. Once you add all the costs together, you will get 17.50. This is cheaper than the day pass, which is 19.95.

zheka24 [161]2 years ago

3 0

I took the test

1. 4

2. 168

3. 2

4. 23.25

5. 54.75

6. 7

7. Pay separately

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Solve the recurrence relation: hn = 5hn−1 − 6hn−2 − 4hn−3 + 8hn−4 with initial values h0 = 0, h1 = 1, h2 = 1, and h3 = 2 using (

musickatia [10]

(a) Suppose $h_n=r^n$ is a solution for this recurrence, with $r\neq0$ . Then

$r^n=5r^{n-1}-6r^{n-2}-4r^{n-3}+8r^{n-4}$
$\implies1=\dfrac5r-\dfrac6{r^2}-\dfrac4{r^3}+\dfrac8{r^4}$
$\implies r^4-5r^3+6r^2+4r-8=0$
$\implies (r-2)^3(r+1)=0\implies r=2,r=-1$

So we expect a general solution of the form

$h_n=c_1(-1)^n+(c_2+c_3n+c_4n^2)2^n$

With $h_0=0,h_1=1,h_2=1,h_3=2$ , we get four equations in four unknowns:

$\begin{cases}c_1+c_2=0\\-c_1+2c_2+2c_3+2c_4=1\\c_1+4c_2+8c_3+16c_4=1\\-c_1+8c_2+24c_3+72c_4=2\end{cases}\implies c_1=-\dfrac8{27},c_2=\dfrac8{27},c_3=\dfrac7{72},c_4=-\dfrac1{24}$

So the particular solution to the recurrence is

$h_n=-\dfrac8{27}(-1)^n+\left(\dfrac8{27}+\dfrac{7n}{72}-\dfrac{n^2}{24}\right)2^n$

(b) Let $G(x)=\displaystyle\sum_{n\ge0}h_nx^n$ be the generating function for $h_n$ . Multiply both sides of the recurrence by $x^n$ and sum over all $n\ge4$ .

$\displaystyle\sum_{n\ge4}h_nx^n=5\sum_{n\ge4}h_{n-1}x^n-6\sum_{n\ge4}h_{n-2}x^n-4\sum_{n\ge4}h_{n-3}x^n+8\sum_{n\ge4}h_{n-4}x^n$
$\displaystyle\sum_{n\ge4}h_nx^n=5x\sum_{n\ge3}h_nx^n-6x^2\sum_{n\ge2}h_nx^n-4x^3\sum_{n\ge1}h_nx^n+8x^4\sum_{n\ge0}h_nx^n$
$G(x)-h_0-h_1x-h_2x^2-h_3x^3=5x(G(x)-h_0-h_1x-h_2x^2)-6x^2(G(x)-h_0-h_1x)-4x^3(G(x)-h_0)+8x^4G(x)$
$G(x)-x-x^2-2x^3=5x(G(x)-x-x^2)-6x^2(G(x)-x)-4x^3G(x)+8x^4G(x)$
$(1-5x+6x^2+4x^3-8x^4)G(x)=x-4x^2+3x^3$
$G(x)=\dfrac{x-4x^2+3x^3}{1-5x+6x^2+4x^3-8x^4}$
$G(x)=\dfrac{17}{108}\dfrac1{1-2x}+\dfrac29\dfrac1{(1-2x)^2}-\dfrac1{12}\dfrac1{(1-2x)^3}-\dfrac8{27}\dfrac1{1+x}$

From here you would write each term as a power series (easy enough, since they're all geometric or derived from a geometric series), combine the series into one, and the solution to the recurrence will be the coefficient of $x^n$ , ideally matching the solution found in part (a).

3 0

2 years ago

QUESTION THREE (30 MARKS) 3.1 The mass of a standard loaf of white bread is, by law meant to be 700g with a population standard

kiruha [24]

Using the normal distribution and the central limit theorem, it is found that there is a 0.0284 = 2.84% probability of finding a sample mean mass of 695g or below.

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Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

----------------------------------

Mean of 700g means that $\mu = 700$
Standard deviation of 21g means that $\sigma = 21$
Sample of 64, thus $n = 64$
For the sampling distribution of the sample mean, the standard deviation is of $s = \frac{21}{\sqrt{64}} = \frac{21}{8} = 2.625$

The probability of finding a sample mean mass of 695g or below is the p-value of Z when X = 695, thus:

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{695 - 700}{2.625}$

$Z = -1.905$

$Z = -1.905$ has a p-value of 0.0284.

0.0284 = 2.84% probability of finding a sample mean mass of 695g or below.

A similar problem is given at brainly.com/question/22934264

7 0

2 years ago

If three sandwiches and two bags of chips cost $22.00, and two sandwiches and one bag of chips cost $14.25, how much does a bag

AfilCa [17]

Answer:

add them up

Step-by-step explanation:

22.00 + 14.25 = 36.25

4 0

1 year ago

A train is leaving in 11 minutes and you are one mile from the station. Assuming you can walk at 4mph and run at 8mph, how much

ddd [48]

7 minutes is the time can you afford to walk

Solution:

Given that, train is leaving in 11 minutes and you are one mile from the station

Let "x" represent how much time can you afford to walk

Then, (11 - x) is the time you run

You can walk at 4mph and run at 8mph

Convert to miles per minute

1 hour = 60 minutes

$4\ mph = 4 \times \frac{1}{60} \text{ miles per minute } = \frac{1}{15} \text{ miles per minute }\\\\8\ mph = 8 \times \frac{1}{60}\ \text{ miles per minute } = \frac{2}{15} \text{ miles per minute }$

We know that,

$Distance = speed \times time$

Given that, you are one mile from the station

Therefore,

1 mile = distance walk + distance run

$1 = x \times \frac{1}{15} + (11-x) \times \frac{2}{15}\\\\1 = \frac{1}{15}(x + 2(11-x))\\\\1 = \frac{1}{15}(x + 22 - 2x)\\\\x + 22 - 2x = 15\\\\x = 22 - 15\\\\x = 7$

Thus, 7 minutes is the time can you afford to walk

6 0

2 years ago

In parallelogram ABCD, the measure of angle ABC is 130, find the measure, in degrees of angle DAB

77julia77 [94]

Angle ABC = 130
Angle ABC = Angle ADC = 130 (Opposite angles are equal)
Angle DAB = 180 - 130 = 50 (consecutive interior angles)

I HOPE IT IS HELPFUL:D

6 0

2 years ago

Read 2 more answers