Ask question

Ask question

All categories

andrezito [222]

2 years ago

14

Thomas consumes coffee ( C) and doughnuts ( D). His budget line was described by the equation D = 20 - 2 C. At a later time, his

budget line could be described by the equation D = 10 - C. The change between the earlier budget line and the later could be explained by the fact that __________________.

1 answer:

dexar [7]2 years ago

4 0

Answer:

Thomas' real income now is half that it was (nominal income decreased by half or prices doubled)

Step-by-step explanation:

If you graph both cases you will see that the shift is parallel between one case and the other.

Changes in relative prices (this is, if only one of the goods change prices), would cause a rotation in the budget line

You might be interested in

Whats 5578 ÷ 68 equal to

kramer

5578/68=82.029411764705882352941176470588

that would be rounded to 82.02 which is the answer

7 0

2 years ago

Read 2 more answers

Clark Heter is an industrial engineer at Lyons Products. He would like to determine whether there are more units produced on the

larisa [96]

Answer:

No, at the 0.05 significance level, the number of units produced on the night shift is not larger.

Step-by-step explanation:

We are given that the mean number of units produced by a sample of 54 day-shift workers was 345. The mean number of units produced by a sample of 60 night-shift workers was 351.

Assume the population standard deviation of the number of units produced on the day shift is 21 and 28 on the night shift.

Let $\mu_1$ = population mean number of units produced on the day shift

$\mu_2$ = population mean number of units produced on the night shift

So, <u>Null Hypothesis</u>, $H_0$ : $\mu_1-\mu_2\geq0$ or $\mu_1\geq\mu_2$ {means that the mean number of units produced on the night shift is same or lesser on the day shift}

<u>Alternate Hypothesis,</u> $H_A$ : $\mu_1-\mu_2$ or $\mu_1$ {means that the mean number of units produced on the night shift is larger}

The test statistics that will be used here is <u>Two-sample z test statistics</u> as we know about population standard deviations;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{\sqrt{\frac{\sigma_1^{2} }{n_1}+\frac{\sigma_2^{2} }{n_2} } }$ ~ N(0,1)

where, $\bar X_1$ = sample mean number of units produced by a sample of 54 day-shift workers = 345

$\bar X_2$ = sample mean number of units produced by a sample of 60 night-shift workers = 351

$\sigma_1$ = population standard deviation of the number of units produced on the day shift = 21

$\sigma_2$ = population standard deviation of the number of units produced on the day shift = 28

$n_1$ = sample of day-shift workers = 54

$n_2$ = sample of night-shift workers = 60

So, <em><u>test statistics</u></em> = $\frac{(345-351)-(0)}{\sqrt{\frac{21^{2} }{54}+\frac{28^{2} }{60} } }$

= -1.302

Now at 0.05 significance level, the z table gives critical value of -1.6449 for left-tailed test. Since our test statistics is more than the critical value of z as -1.302 > -1.6449 so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that the mean number of units produced on the night shift is same or lesser than those produced on the day shift.

4 0

2 years ago

hjlf

Answer:

240

Step-by-step explanation:

4 0

2 years ago

Determine whether the series is convergent or divergent. 1 2 3 4 1 8 3 16 1 32 3 64 convergent divergent Correct:

Vanyuwa [196]

Answer:

This series diverges.

Step-by-step explanation:

In order for the series to converge, i.e. $\lim_{n \to \infty} a_n =A$ it must hold that for any small $\epsilon$ >0, there must exist $n_0\in \mathbb{N}$ so that starting from that term of the series all of the following terms satisfy that $|a_n-A|n_0$ .

It is obvious that this cannot hold in our case because we have three sub-series of this observed series. One of them is a constant series with $a_n=1$ , the other is constant with $a_n=3$ , and the third one has terms that are approaching infinity.

Really, we can write this series like this:

$a_n=\begin{cases} 1 \ , \ n=4k+1, k\in \mathbb{N}_0\\ 2^{k}\ , \ n=2k, k\in \mathbb{N}_0\\3\ , \ n=4k+3, k\in \mathbb{N}_0\end{cases}$

If we denote the first series as $b_n=1$ , we will have that $\lim_{k \to \infty} b_k=1$ .

The second series is denoted as $c_k=2^k$ and we have that $\lim_{k \to \infty} c_k=+\infty$ .

The third sub-series $d_k=3$ is a constant series and it holds that $\lim_{k \to \infty} d_k=3$ .

Since those limits of sub-series are different, we can never find such $n_0\\$ so that every next term of the entire series is close to one number.

To make an example, if we observe the first sub-series if follows that A must be equal to 1. But if we chose $\epsilon =1$ , all those terms associated with the third sub-series will be out of this interval $(A-1, A+1)=(0, 2)$ .

Therefore, the observed series diverges.

5 0

2 years ago

One of the factors of the polynomial x3 − 5x2 is x + 3. What is the other factor?

blondinia [14]

Answer:

C. x² − 8x + 24 − 72/(x+3)

Step-by-step explanation:

See attached picture for long division method.

Logically, we know x³ − 5x² factors to x² (x − 5). Since x + 3 isn't a factor, we know the remainder isn't 0. So we can narrow the options down to A or C.

One way to find the remainder is through long division. Or, since this is multiple choice, we multiply the options by x + 3 and see which one results in an answer of x³ − 5x².

(x + 3) (x² − 8x + 24 − 72/(x+3))

(x + 3) (x² − 8x + 24) − 72

x³ − 8x² + 24x + 3x² − 24x + 72 − 72

x³ − 5x²

6 0

2 years ago