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3 years ago

Over the course of 4 days the price of stock at a corporation fell $28. What was the average change in stock price per day?

Mathematics

1 answer:

bezimeni [28]3 years ago

5 0

7 is the average stock change

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Would a yardstick be reasonable to measure the length of a canoe paddle

pishuonlain [190]

Most charts for the sizes of the canoe paddle give the sizes in units of inches. A yard is a unit of measure of length that is equal to 36 inches. Although some sizes of the canoes paddle torso are not greater than 36 inches, the yardstick can still be a reasonable measuring tool for bigger canoe paddles.

8 0

4 years ago

Tristan had 95 inches of ribbon.He cut off 27.50 inches. How many inches of ribbon remain

tatiyna

Answer:

67.50 inches

Step-by-step explanation:

5 0

4 years ago

-8(8v + 1) - 2 = -394

Bess [88]

Answer:

v = 6

Step-by-step explanation:

Solve for v:

-8 (8 v + 1) - 2 = -394

-8 (8 v + 1) = -64 v - 8:

-64 v - 8 - 2 = -394

Grouping like terms, -64 v - 8 - 2 = -64 v + (-8 - 2):

-64 v + (-8 - 2) = -394

-8 - 2 = -10:

-10 - 64 v = -394

Add 10 to both sides:

(10 - 10) - 64 v = 10 - 394

10 - 10 = 0:

-64 v = 10 - 394

10 - 394 = -384:

-64 v = -384

Divide both sides of -64 v = -384 by -64:

(-64 v)/(-64) = (-384)/(-64)

(-64)/(-64) = 1:

v = (-384)/(-64)

The gcd of -384 and -64 is -64, so (-384)/(-64) = (-64×6)/(-64×1) = (-64)/(-64)×6 = 6:

Answer: v = 6

3 0

3 years ago

Read 2 more answers

How many Solutions does this system have? (1 point)

mixas84 [53]

The given system of equation that is $2x+y=3$ and $6x=9-3y$ has infinite number of solutions.

Option -C.

<u>Solution:</u>

Need to determine number of solution given system of equation has.

$\begin{array}{l}{2 x+y=3} \\\\ {6 x=9-3 y}\end{array}$

Let us first bring the equation in standard form for comparison

$\begin{array}{l}{2 x+y-3=0} \\\\ {6 x+3 y-9=0}\end{array}$

$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

To check how many solutions are there for system of equations $a_{1} x+b_{1} y+c_{1}=0 \text{ and }a_{2} x+b_{2} y+c_{2}=0$ , we need to compare ratios of $\frac{a_{1}}{a_{2}}, \frac{b_{1}}{b_{2}} \text { and } \frac{c_{1}}{c_{2}}$

In our case,

$a_{1} = 2, b_{1}= 1\text{ and }c_{1}= -3$

$a_{2} = 6, b_{2} = 3,\text{ and }c_{2} = -9$

$\begin{array}{l}{\Rightarrow \frac{a_{1}}{a_{2}}=\frac{2}{6}=\frac{1}{3}} \\\\ {\Rightarrow \frac{b_{1}}{b_{2}}=\frac{1}{3}} \\\\ {\Rightarrow \frac{c_{1}}{c_{2}}=\frac{-3}{-9}=\frac{1}{3}} \\\\ {\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}=\frac{1}{3}}\end{array}$

As $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ , so given system of equations have infinite number of solutions.

Hence, we can conclude that system has infinite number of solutions.

5 0

3 years ago

According to the U.S. Bureau of Labor Statistics, 20% of all people 16 years of age or older do volunteer work. In this age grou

Murljashka [212]

Answer:

1. P(X≥35) = 0.0183

2. P(X≤21) = 0.0183

3. P(0.18<p<0.25) = 0.7915

Step-by-step explanation:

We have the proportion for women: pw=0.22, and the proportion for men: pm=0.19.

1. We have a sample of 140 woman and we have to calculate the probability of getting 35 or more who do volunteer work.

This is equivalent to a proportion of

$p=X/n=35/140=0.25$

The standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.22*0.78}{300}}\\\\\\ \sigma_p=\sqrt{0.0006}=0.0239$

We calculate the z-score as:

$z=\dfrac{p-p_w}{\sigma_p}=\dfrac{0.25-0.22}{0.0239}=\dfrac{0.03}{0.0239}=0.8198$

Then, the probability of having 35 women or more who do volunteer work in this sample of 140 women is:

$P(X>35)=P(p>0.25)=P(z>2.0906)=0.0183$

2. We have to calculate the probability of having 21 or fewer women in the group who do volunteer work.

The proportion is now:

$p=X/n=21/140=0.15$

We can calculate then the z-score as:

$z=\dfrac{p-p_w}{\sigma_p}=\dfrac{0.15-0.2}{0.0239}=\dfrac{-0.05}{0.0239}=-2.0906$

Then, the probability of having 21 women or less who do volunteer work in this sample of 140 women is:

$P(X$

3. For the sample with men and women, we use the proportion for both, which is π=0.2.

The sample size is n=300.

Then, the standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.2*0.8}{300}}\\\\\\ \sigma_p=\sqrt{0.0005}=0.0231$

We can calculate the z-scores for p1=0.18 and p2=0.25:

$z_1=\dfrac{p_1-\pi}{\sigma_p}=\dfrac{0.18-0.2}{0.0231}=\dfrac{-0.02}{0.0231}=-0.8660\\\\\\z_2=\dfrac{p_2-\pi}{\sigma_p}=\dfrac{0.25-0.2}{0.0231}=\dfrac{0.05}{0.0231}=2.1651$

We can now calculate the probabilty of having a proportion within 0.18 and 0.25 as:

$P=P(0.18$

5 0

4 years ago