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I am Lyosha
1 month ago
7

The Baum and the Freeman families are comparing their electric bills for the past month. The Baum family is on a standard use pl

an and the Freeman family is on an interval use plan. Each family’s usage is listed on the chart below:
Standard Use Plan
Interval Use Plan
8.5 cents/kWh for the first 400 kWh
12 cents/kWh for the next 400 kWh
14.5 cents/kWh for anything over 800 kWh
On-peak hours - 13 cents/kWh
Off-peak hours - 2 cents/kWh

Both families use 1250 kWh for the given 30 day period. The Freeman family uses 400 kWh during on-peak hours and 850 during off-peak hours. Which family ends up paying more for their utilities? How much more?
a.
The Freeman family pays $260.83 more than the Baum family.
b.
The Freeman family pays $250.00 more than the Baum family.
c.
The Baum family pays $69.00 more than the Freeman family.
d.
The Baum family pays $78.25 more than the Freeman family.
Mathematics
2 answers:
tester [12.3K]1 month ago
5 0
La opción D es la correcta. Solo respondí 100 en mi examen y ¡yoyo!
Svet_ta [12.7K]1 month ago
4 0
La familia Freeman:
13 c/kWh (400 kWh) + 2 c/kWh (400 kWh) + 14.5 c/kWh) 450 kWh
 = 6464.5 c
6464.5 c x 30 = 193935

La familia Baum:
8.5 (400) + 12 (400) + 14.5 (450)
= 14725 c

<span>La familia Freeman tiene un costo de $260.83 más que la familia Baum.
</span>
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In a circle, an arc 60 centimeters long subtends an angle of 5 radians. What is the radius of the circle?
Inessa [12570]
Got it!
There are 2π radians in a complete circle.

Now, let's calculate the circumference.
5/2π = 60/circumference.
Next, solve for the circumference.
By multiplying both sides by 2π, we have: 5 * circumference = 120π.
Now divide both sides by 5, and we find: circumference = 24π.

Using the formula c = 2πr,
we set 24π = 2πr.
Dividing both sides by 2π gives us r = 12. Thus, the radius measures 12cm.
6 0
3 months ago
Read 2 more answers
A pile of sand has a weight of 90kg the sand is put into a small bag and a medium bag and a large bag in the ratio 2:3:7
tester [12383]
90kg of sand was divided into 3 bags, constituting a total of 2+3+7=12 sections.
90kg/12=7.5kg
The smaller bag comprises 2 sections, hence it weighs 7.5kg*2=15kg
The medium-sized bag consists of 3 sections, leading to a total weight of 7.5kg*3=22.5kg
The larger bag encompasses 7 sections, resulting in a weight of 7.5kg*7=52.5kg
Thus, the ratio 2:3:7 translates to 15kg:22.5kg:52.5kg
Verifying, 15kg+22.5kg+52.5kg equals 90kg
3 0
2 months ago
Robby had 4 4/9 bags of pet food.All of the bags held the same amount of food when they were full.Of Robby's 4 4/9 bags of pet f
Zina [12379]

Respuesta: 1.8 bolsas

Explicación paso a paso:

A partir de la pregunta, considerando que Robby tiene 4 4/9 bolsas de comida para mascotas, 2/5 eran comida para perros.

2/5 de 4 4/9 = comida para perros.

Convierte 4 4/9 a fracción impropia= 40/9.

2/5 de 40/9 implica 2/5 × 40/9.

= 16/9.

= 1.78 o 1.8.

Espero que esto sea útil, por favor marca como la mejor respuesta.

4 0
3 months ago
There were 5,317 previously owned homes sold in a western city in the year 2000. The distribution of the sales prices of these h
tester [12383]

Answer:

(A) Approximately normal with a mean of $206,274 and a standard deviation of $3,788.

Step-by-step explanation:

The Central Limit Theorem asserts that for a random variable X that follows a normal distribution with a mean of \mu and a standard deviation of \sigma, the sampling distribution of sample means, when drawn with size n, can be estimated as a normal distribution with a mean of \mu and a standard deviation of s = \frac{\sigma}{\sqrt{n}}.

Even if the variable is skewed, as long as n is no less than 30, the Central Limit Theorem still holds.

Population:

Right skewed

Mean $206,274

Standard deviation $37,881.

Sample:

<pbased on="" the="" central="" limit="" theorem="" it="" can="" be="" approximated="" to="" normal.="">

Mean $206,274

Standard deviation s = \frac{37881}{\sqrt{100}} = 3788.1

So the correct answer is:

(A) Approximately normal with a mean of $206,274 and a standard deviation of $3,788.

</pbased>
4 0
2 months ago
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