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2 years ago

Anatoly graphed his store’s costs below. Which statement best describes the rate of change in anatomy’s graph?

Mathematics

1 answer:

GarryVolchara [31]2 years ago

5 0

Answer:

Ansswer: C. Anatoly's costs decrease by $50 for every unit sold

Step-by-step explanation:

In this graph Cost (in $100) has been plotted on Y axis and Units sold on X axis. It is evident from the graph that at zero sale per unit cost was $1600 which reduced down to $1400 when 4 units were sold.

Therefore from $1600 to $1400 cost was reduced by $200 and 4 units were sold.

So reduction in cost per unit will be = $200/4 = $ 50

which is the final answer.

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Miyoko's goal is to earn more than \$950$950dollar sign, 950 this week. She earns \$250$250dollar sign, 250 for every day she wo

GrogVix [38]

$250 c+ $ 180 g > $ 950

<u>Step-by-step explanation:</u>

As a cryptographer (c), Miyoko earns per day = $ 250

As a geologist (g) , Miyoko earns per day = $ 180

So the equation comes to be $250 c+ $ 180 g = $ 950

The equation can be rewritten to find c as, (950-180 g) / 250

The equation can be rewritten to find g as, (950 - 250 c) / 180

Plugin different values of c and g in the above 2 equations, we can find that ,

To achieve the goal, Miyoko requires to be a geologist for 3 days and crpytographist for 2 days.

5 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

Elena bikes 20 minutes each day for exercise. Write an equation to describe the relationship between her distance in miles, D, a

ValentinkaMS [17]

Answer:

The equation to describe the relationship between Elena distance:

a) D = 4.34 miles, b) D = 4.25 miles and c) 12D = M + 3N miles.

Solution:

Given, Elena bikes 20 minutes each day for exercise.

We have to write an equation to describe the relationship between her distance in miles, D, and her biking speed, in miles per hour,

We know that, distance travelled = speed time

a. At a constant speed of 13 miles per hour for the entire 20 minutes

Her speed is 13 miles per hour.

Then, distance D miles = 13 miles per hour 20 minutes

b. At a constant speed of 15 minutes per hour for the first 5 minutes, then at 12 miles per hour for the last 15 minutes

Now, total distance travelled = distance travelled with 15 mph + distance travelled with 12 mph

c. At a constant speed of M miles per hour for the first 5 minutes, then at N miles per hour for the last 15 minutes

Now, total distance travelled = distance travelled with M mph + distance travelled with N mph

Hence, a) D = 4.34 miles, b) D = 4.25 miles and c) 12D = M + 3N miles.

4 0

2 years ago

Select all the correct graphs.<br> Choose the graphs that indicate equations with no solution

galben [10]

Answer:

You didnt post any graphs, also the no solution graphs are graphs that have lines that never touch

Step-by-step explanation:

7 0

2 years ago

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MrRissso [65]

I pound of carrots would cost 13 pounds

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2 years ago