Ask question

Ask question

All categories

3 years ago

10

The price of a stock decrease by 60 cents one week , decreased 10 cents the next week , and decreased another 20 cents the follo

wing week
What is the average change in the price of the stock over the three weeks ?

-270 cents

-90

-87

-30

2 answers:

Maru [420]3 years ago

8 0

It's -90. I'm positive.<span />

r-ruslan [8.4K]3 years ago

7 0

-60 + -10 + -20 = -90 / 3 = -30

You might be interested in

Brittany buys 2.55 pounds of turkey for $5.96 per pound and 3.7 pounds of

Galina-37 [17]

Answer:

Brittany needs another $3.7405.

Step-by-step explanation:

Per pound Cost of turkey = $5.96 per pound

The amount Brittany buys the turkey = 2.55 pounds

Brittany's cost for turkey = 2.55 × $5.96 = $15.198

Per pound cost for cheese = $3.35 per pound

The amount Brittany buys the cheese = 3.7 pounds

Brittany's cost for cheese = 2.55 × $3.35 = $8.5425

So,

Brittany's total cost = Turkey cost + Cheese cost

= $15.198 + $8.5425

= $23.7405

As brittany gave the clerk 20 dollars.

So, the amount she further needs will be:

$23.7405 - $20 = $3.7405

Therefore, Brittany needs another $3.7405.

8 0

2 years ago

Suppose we are interested in analyzing the weights of NFL players. We know that on average, NFL players weigh 247 pounds with a

tigry1 [53]

Answer:

$SE= \frac{\sigma}{\sqrt{n}} = \frac{47}{\sqrt{30}}= 8.58$

$ME= 1.64 *\frac{\sigma}{\sqrt{n}} = 1.64*\frac{47}{\sqrt{30}}= 14.073$

$\bar X - ME = 237- 14.073 = 222.927$

$\bar X + ME = 237+ 14.073 = 251.073$

$n=(\frac{1.640(47)}{5})^2 =237.65 \approx 238$

So the answer for this case would be n=238 rounded up to the nearest integer

Null hypothesis: $\mu \geq 247$

Alternative hypothesis: $\mu$

$z=\frac{237-247}{\frac{47}{\sqrt{30}}}=-1.165$

$p_v =P(z$

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't conclude that the true mean is lower than 247 at 10% of significance.

Step-by-step explanation:

For this case we have the following data given:

$\bar X =237$ represent the sample mean

$\sigma = 47$ represent the population deviation

$n =30$ represent the sample size selected

$\mu_0 = 247$ represent the value that we want to test.

The standard error for this case is given by:

$SE= \frac{\sigma}{\sqrt{n}} = \frac{47}{\sqrt{30}}= 8.58$

For the 90% confidence the value of the significance is given by $\alpha=1-0.9 = 0.1$ and $\alpha/2 = 0.05$ so we can find in the normal standard distribution a quantile that accumulates 0.05 of the area on each tail and we got:

$z_{\alpha/2}= 1.64$

And the margin of error would be:

$ME= 1.64 *\frac{\sigma}{\sqrt{n}} = 1.64*\frac{47}{\sqrt{30}}= 14.073$

The confidence interval for this case would be given by:

$\bar X - ME = 237- 14.073 = 222.927$

$\bar X + ME = 237+ 14.073 = 251.073$

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

And on this case we have that ME =5 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

Replacing into formula (b) we got:

$n=(\frac{1.640(47)}{5})^2 =237.65 \approx 238$

So the answer for this case would be n=238 rounded up to the nearest integer

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean is lower than 247 pounds, the system of hypothesis would be:

Null hypothesis: $\mu \geq 247$

Alternative hypothesis: $\mu$

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{237-247}{\frac{47}{\sqrt{30}}}=-1.165$

P-value

Since is a left tailed test the p value would be:

$p_v =P(z$

Conclusion

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't conclude that the true mean is lower than 247 at 10% of significance.

5 0

2 years ago

7x - 4 = 24<br> show your work

Debora [2.8K]

Move -4 to the other side
7x=24+4
7x=28
x=4

4 0

3 years ago

Read 2 more answers

If you’re good with geometry can you help me?

Paladinen [302]

Answer:

a: AB is perpendicular to CD

b: AM is equal in length to BM

4 0

3 years ago

Please give answer of 6 number

Irina18 [472]

Answer:

$7,500$

Step-by-step explanation:

Let's solve this problem step-by-step. The library had 1,500 books in 2011. The ratio of books in 2011 and in 2012 is 1:2. Therefore, let the number of books in 2012 be $x$ .

We have the following proportion:

$\frac{1}{1,500}=\frac{2}{x},\\x=2\cdot 1,500=3,000$

Therefore, there were 3,000 books in 2012. The ratio of books in 2012 and in 2013 is 2:5. Let the number of books in 2013 be $y$ .

We have:

$\frac{2}{3,000}=\frac{5}{y},\\2y=5\cdot 3,000,\\2y=15,000\\y=\boxed{7,500}$

Therefore, there were 7,500 books in 2013.

8 0

3 years ago

Read 2 more answers