How to solve word problems wit fractions

Choose the situation in which using exact numbers is the MOST appropriate.

QveST [7]

Answer:

A, you need to pay with exact change

3 0

3 years ago

Read 2 more answers

The circle below is centered at the point (4,3) and has a radius length 5. What is its equation

Crazy boy [7]

The equation of a circle centred at point (m,n) and radius r is given by

<span>(x-m)² + (y-n)² = r²
</span>-------------------------------------------------------------
Centre = (4,3)
radius = 5

Equation:
(x - 4)² + (y - 3)² = 5²
⇒ x² - 8x + 16 + y² - 6y + 9 = 25
⇒ x² + y² - 8x - 6y + 25 = 25
⇒ x² + y² - 8x - 6y = 0

The equation of the circle is x² + y² - 8x - 6y = 0

Hope it helps!

3 0

3 years ago

Read 2 more answers

Write a paragraph comparing the two classes' semester grades. Be sure to compare the extremes, the quartiles, the medians, and t

Gnom [1K]

Answer:

Second class have higher marks and greater spread.

Step-by-step explanation:

First box plot represents class first. From the first box plot, we get

$\text{Minimum value }= 53,Q_1=62,Median=80,Q_3=86,\text{Maximum value }= 89$

$Range=Maximum-Minimum=89-53=36$

$IQR=Q_3-Q-1=86-62=24$

Second box plot represents class second. From the second box plot, we get

$\text{Minimum value }= 56,Q_1=62,Median=74,Q_3=89,\text{Maximum value }= 96$

$Range=Maximum-Minimum=96-56=40$

$IQR=Q_3-Q-1=89-62=27$

First class has greater minimum value, first quartile of both classes are same, second class has greater median, first class has greater third quartile and first class has greater maximum value. It means second class have higher marks but class first have less variation.

Second class has greater range and greater inter quartile range. It means data of second class has greater spread.

Therefore, second class have higher marks and greater spread.

3 0

3 years ago

Carla needs 8 inches of ribbon for each craft she makes. What is the greatest number of crafts Carla can make using 30 feet of r

Contact [7]

Well 30x12in is 360in. 360in divided by 8 is 45. She can make 45 crafts

7 0

3 years ago

In 2010, the world's largest pumpkin weighed 1,810 kilograms. An average-sized pumpkin weighs 5,000 grams.

jekas [21]

Answer:

1,805 kg.

Step-by-step explanation:

We have been given that in 2010, the world's largest pumpkin weighed 1,810 kilograms. An average-sized pumpkin weighs 5,000 grams. We are asked to find the how much the world's largest pumpkin weighs than an average pumpkin.

First of all, we will convert the weight of average pumpkin in kilograms by dividing 5,000 by 1000 as 1 kg equals 1,000 gm.

$\text{The weight of average pumpkin}=\frac{5000\text{ gm}}{\frac{\text{1000 gm}}{\text{ 1 kg}}}$

$\text{The weight of average pumpkin}=5000\text{ gm}\times \frac{\text{ 1 kg}}{\text{1000 gm}}$

$\text{The weight of average pumpkin}=5\text{ kg}$

Now, we will subtract the weight of average pumpkin from world's largest pumpkin's weight.

$1,810\text{ kilograms}-5\text{ kilograms}=1,805\text{ kilograms}$

Therefore, the 2010 world-record pumpkin weighs 1,805 kilograms more than an average-sized pumpkin.

3 0

4 years ago

Read 2 more answers