People in a town voted for a new councillor.

Dividends Per Share Seventy-Two Inc., a developer of radiology equipment, has stock outstanding as follows: 60,000 shares of cum

anygoal [31]

Answer and Step-by-step explanation:

The computation of dividends per share on each class of stock for each of the four years is shown below:-

Particulars 1st year 2nd-year 3rd-year 4th year

Preferred dividend

paid a $34,000 $38,000 $36,000 $36,000

Number of preferred

stock b 60,000 60,000 60,000 60,000

Dividend per share

(a ÷ b) $0.57 $0.63 $0.60 $0.60

Dividend paid to common

stockholders c $0 $38,000 $44,000 $64,000

Number of common stock

shares d 410,000 410,000 410,000 410,000

Dividend per share

on common stock $0 $0.093 $0.11 $0.16

(c ÷ d)

Working note:

Preferred dividend = Number of preferred stock shares × Par value per share × Percentage of dividend

= 60,000 × $20 × 3%

= $36,000

Preferred stock

For 1st year

= $34,000

For 2nd-year

Dividend in year 2+ Dividend balance in year 1

= $36,000 + ($36,000 - $34,000)

= $38,000

For 3rd-year

= $36,000

For 4th year

= $36,000

Common stock dividend

Particulars 1 year 2 year 3 year 4 year

Total dividend paid $34,000 $76,000 $80,000 $100,000

Less:

Preferred stock

dividend $34,000 $38,000 $36,000 $36,000

Dividend paid to common

stockholders $0 $38,000 $44,000 $64,000

8 0

2 years ago

There are 12 boys and 8 girls in a class, including a brother and a sister. If one boy and one girl are selected at random from

kupik [55]

I think the answer is 1/10.

7 0

2 years ago

The pucks used by the National Hockey League for ice hockey must weigh between and ounces. Suppose the weights of pucks produced

Dahasolnce [82]

Answer:

$P(5.5$

And we can find this probability using the normal standard distribution or excel and we got:

$P(-2.769$

Step-by-step explanation:

For this case we assume the following complete question: "The pucks used by the National Hockey League for ice hockey must weigh between 5.5 and 6 ounces. Suppose the weights of pucks produced at a factory are normally distributed with a mean of 5.86 ounces and a standard deviation of 0.13ounces. What percentage of the pucks produced at this factory cannot be used by the National Hockey League? Round your answer to two decimal places. "

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

$X \sim N(5.86,0.13)$

Where $\mu=5.86$ and $\sigma=0.13$

We are interested on this probability

$P(5.5$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(5.5$

And we can find this probability using the normal standard distribution or excel and we got:

$P(-2.769$

4 0

2 years ago

A brewery produces cans of beer that are supposed to contain exactly 12 ounces. But owing to the inevitable variation in the fil

MArishka [77]

Answer:

$T \sim N (\mu = 6*12=72 , \sigma= \sqrt{6} *0.3=0.735)$

$P(T \leq 72) = P(Z< \frac{72-72}{0.735}) = P(Z$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solutio to the problem

Let X the random variable that represent the amount of beer in each can of a population, and for this case we know the distribution for X is given by:

$X \sim N(12,0.3)$

Where $\mu=12$ and $\sigma=0.3$

For this case we select 6 cans and we are interested in the probability that the total would be less or equal than 72 ounces. So we need to find a distribution for the total.

The definition of sample mean is given by:

$\bar X = \frac{\sum_{i=1}^n X_i}{n} = \frac{T}{n}$

If we solve for the total T we got:

$T= n \bar X$

For this case then the expected value and variance are given by:

$E(T) = n E(\bar X) =n \mu$

$Var(T) = n^2 Var(\bar X)= n^2 \frac{\sigma^2}{n}= n \sigma^2$

And the deviation is just:

$Sd(T) = \sqrt{n} \sigma$

So then the distribution for the total would be also normal and given by:

$T \sim N (\mu = 6*12=72 , \sigma= \sqrt{6} *0.3=0.735)$

And we want this probability:

$P(T\leq 72)$

And we can use the z score formula given by:

$z = \frac{x-\mu}{\sigma}$

$P(T \leq 72) = P(Z< \frac{72-72}{0.735}) = P(Z$

6 0

2 years ago

According to a Los Angeles Times study of more than 1 million medical dispatches from 2007 to 2012, the 911 response time for me

Reil [10]

Answer:

a) $\bar X=10.65$

$Median =\frac{10.7+10.7}{2}=10.7$

$Mode= 10.7$

b) $Range = 11.8-8.3=3.5$

$s= 0.948$

c) $IQR = Q_3 -Q_1 = 11.05-10.55=0.5$

And we can find the usual limits with:

$Lower = Q_1 -1.5 IQR = 10.55 -1.5*0.5=9.8$

$Upperer = Q_3 +1.5 IQR = 10.55 +1.5*0.5=11.3$

And since 8.3 <9.8 we can consider this value too low or as an outlier for this case.

d) The mean for this case was 10.65 and the usual values are between 9.8 and 11.3, so as we can see all are above the specified value of 6 minutes, and we can conclude that the times are not satisfying the quality standards for this case.

And they should be considered apply some strategies to reduce the response time, adding more stations around points selected at the city could be useful in order to reduce the response time.

Step-by-step explanation:

We have the following data:

11.8 10.3 10.7 10.6 11.5 8.3 10.5 10.9 10.7 11.2

Part a

We can calculate the sample mean with the following formula:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And if we replace we got: $\bar X=10.65$

For the median we need to sort the values on increasing order and we have:

8.3 10.3 10.5 10.6 10.7 10.7 10.9 11.2 11.5 11.8

Since n =10 we can calculate the median as the average between the 5th and 6th position of the dataset ordered and we got:

$Median =\frac{10.7+10.7}{2}=10.7$

The mode would be the most repeated value on this case:

$Mode= 10.7$

Part b

The range is defined as $Range =Max-Min$ and if we replace we got:

$Range = 11.8-8.3=3.5$

We can calculate the standard deviation with the following formula:

$s= sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And if we replace we got:

$s= 0.948$

Part c

For this case we can use the IQR method in order to determine if 8.3 is an outlier or not.

We can calculate the first quartile with these values: 8.3 10.3 10.5 10.6 10.7 10.7 and $Q_1= \frac{10.6+10.7}{2}=10.55$

And for the Q3 we can use: 10.7 10.7 10.9 11.2 11.5 11.8 and we got $Q_3 = \frac{10.9+11.2}{2}=11.05$

Then we can find the IQR like this:

$IQR = Q_3 -Q_1 = 11.05-10.55=0.5$

And we can find the usual limits with:

$Lower = Q_1 -1.5 IQR = 10.55 -1.5*0.5=9.8$

$Upperer = Q_3 +1.5 IQR = 10.55 +1.5*0.5=11.3$

And since 8.3 <9.8 we can consider this value too low or as an outlier for this case.

Part d

The mean for this case was 10.65 and the usual values are between 9.8 and 11.3, so as we can see all are above the specified value of 6 minutes, and we can conclude that the times are not satisfying the quality standards for this case.

And they should be considered apply some strategies to reduce the response time, adding more stations around points selected at the city could be useful in order to reduce the response time.

6 0

2 years ago