All categories

2 years ago

How long does it take Natsumi to clean up after she is done?

Mathematics

2 answers:

Snezhnost [94]2 years ago

8 0

Step-by-step explanation:

Vaselesa [24]2 years ago

7 0

Answer:

<u>0.5 hours</u>

Step-by-step explanation:

Let the total time to paint and clean = y

and x is the area Natsumi paints

As she paints with a constant speed, so, there is a linear variation between x and y and then she takes a constant a mount to clean

So, the equation which describe this situation is ⇒ y = ax + b

Where a = hours/area and b is a constant a mount to clean.

Using the given table:

at x = 30 m² ⇒ y = 2 hours ⇒ 2 = 30 a + b ⇒ (1)

at x = 45 m² ⇒ y = 2.75 hours ⇒ 2.75 = 45 a + b ⇒(2)

solve (1) and (2) to find a and b

from (1) ⇒ b = 2 - 30a ⇒ (3)

Substitution with (3) at (2)

2.75 = 45a + 2 - 30a

Solve for a:

2.75 - 2 = 45a - 30a

0.75 = 15a

a = 0.75/15 = 0.05 hours/m²

b = 2 - 30a = 2 - 30*0.05 = 0.5 hours

Check the results using the third raw:

at x = 60

∴ y = 0.05 * 60 + 0.5 = 3.5 hours (correct as in table)

So, the time taken to clean up after she is done = 0.5 hours

You might be interested in

The CEO of a local tech company wants to estimate the proportion of employees that are pleased with the cleanliness of the break

Zigmanuir [339]

Answer:

Methods of obtaining a sample of 600 employees from the 4,700 workforce:

Part A: The type of sampling method proposed by the CEO is Convenience Sampling.

Part B: When there are equal number of participants in both campuses, stratification by campus would give a more precise approximation of the proportion of employees who are satisfied with the cleanliness of the breakrooms than stratification by gender. Another method to ensure that stratification by campus gives a more precise approximation of the proportion of employees who are satisfied with the cleanliness of the breakrooms than stratification by gender is to ensure that the sample is proportional to the proportion of each campus to the whole population or workforce.

Step-by-step explanation:

A Convenience Sampling technique is a non-probability (non-random) sampling method and the participants are selected based on availability (early attendees). The early attendees might be different from the late attendees in characteristics such as age, sex, etc. Therefore, sampling biases are present. All non-probability sampling methods are prone to volunteer bias.

Stratified sampling is more accurate and representative of the population. It reduces sampling bias. The difficulty arises in choosing the characteristic to stratify by.

3 0

2 years ago

Hillel is juggling flaming torches to raise money for charity. His initial appearance raises \$500$500dollar sign, 500, and he r

Svetlanka [38]

Answer:

$R = $500 + $15(t)

Step-by-step explanation:

in this question, we are told to give a mathematical expression that would model the amount of money that is made by Hillel.

from the question, we were made to understand that the amount of money he makes is a front of his time t. that is noted.

and also, he has a base cost of $500.

now, let’s write an equation;

R = $500 + 15(t)

where t represents the length of the performance as suggested by women

3 0

2 years ago

Malcolm and Ravi raced each other.

Semenov [28]

Malcolm maximum speed is 200 km/h and Ravi maximum speed is 320 km/h

Step-by-step explanation:

Lets assume Malcolm maximum speed to be M km/h and Ravi maximum speed to be R km/h

Average of their maximum speed is; (M+R)/2 =260 km/h

This can be simplified to M+R= 260*2

M+R = 520.......................................(I)

When Malcolm's speed is doubled,

2M=R+80 km/h ----------------------(ii)

The two equations are;

M+R = 520 ------ M=520-R ----(i)

2M=R+80 ---------------------------(ii)

Replacing M with (i) above

2M=R+80

2(520-R) =R+80

1040 -2R =R+80

1040-80 = R+2R

960 = 3R

960/3 =R

320 =R

M+R =520

M=520-320 =200

Malcolm maximum speed is 200 km/h and Ravi maximum speed is 320 km/h

Learn More

Simultaneous equations:brainly.com/question/12919422

Keywords: Average, maximum, speeds, doubled,

#LearnwithBrainly

3 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

Emily and a friend bought two tickets to see a soccer game. Each ticket cost $8.25. The friends paid a total of $24.50, which in

solniwko [45]

$2\cdot8.25=16.5\\\\24.5-16.5=8\\\\8:2=4\\\\\frac{4}{8.25}\cdot100\%=\\\\\frac{400}{8.25}\%=\\\\48.48\%$

6 0

2 years ago