$79.95 rounded to the nearest cent

Garth buys 24 kilograms of candy. He then repacks the candy into small bags of 34 kilogram each. How many bags does he pack? A.

LenaWriter [7]

<u>Corrected Question</u>

Garth buys 24 kilograms of candy. He then repacks the candy into small bags of 3/4 kilogram each. How many bags does he pack?

Answer:

C. 32 bags

Step-by-step explanation:

Total Weight of Bag bought =24 kilograms

Since Garth wants to repack the candy into small bags of 3/4 kg, we simply divide 24 by 3/4 to obtain the number of small bags.

$24 \div \dfrac{3}{4}= 24 X \dfrac{4}{3}\\=32$

There will be 32 small bags of 3/4 kg.

The correct option is C.

6 0

3 years ago

There is a line through the origin that divides the region bounded by the parabola y=2x-4x^2 and the x-axis into two regions wit

shtirl [24]

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

y = 7x - 4x²

<span>7x - 4x² = 0 </span>

<span>x(7 - 4x) = 0 </span>

<span>x = 0, 7/4 </span>

<span>Find the area of the bounded region... </span>

<span>A = ∫ 7x - 4x² dx |(0 to 7/4) </span>

<span>A = 7/2 x² - 4/3 x³ |(0 to 7/4) </span>

<span>A = 7/2(7/4)² - 4/3(7/4)³ - 0 = 3.573 </span>

<span>Half of this area is 1.786, now set up an integral that is equal to this area but bounded by the parabola and the line going through the origin... </span>

<span>y = mx + c </span>

<span>c = 0 since it goes through the origin </span>

<span>The point where the line intersects the parabola we shall call (a, b) </span>

<span>y = mx ===> b = m(a) </span>

<span>Slope = m = b/a </span>

<span>Now we need to integrate from 0 to a to find the area bounded by the parabola and the line... </span>

<span>1.786 = ∫ 7x - 4x² - (b/a)x dx |(0 to a) </span>

<span>1.786 = (7/2)x² - (4/3)x³ - (b/2a)x² |(0 to a) </span>

<span>1.786 = (7/2)a² - (4/3)a³ - (b/2a)a² - 0 </span>

<span>1.786 = (7/2)a² - (4/3)a³ - b(a/2) </span>

<span>Remember that (a, b) is also a point on the parabola so y = 7x - 4x² ==> b = 7a - 4a² </span>
<span>Substitute... </span>

<span>1.786 = (7/2)a² - (4/3)a³ - (7a - 4a²)(a/2) </span>

<span>1.786 = (7/2)a² - (4/3)a³ - (7/2)a² + 2a³ </span>

<span>(2/3)a³ = 1.786 </span>

<span>a = ∛[(3/2)(1.786)] </span>

<span>a = 1.39 </span>

<span>b = 7(1.39) - 4(1.39)² = 2.00 </span>

<span>Slope = m = b/a = 2.00 / 1.39 = 1.44</span>

7 0

3 years ago

Last Sunday a certain store sold copies of Newspaper A for $1 and Newspaper B for 1.25. If r percent of the stores revenue came

Oksanka [162]

Answer:

Option 400p/500-p

Step-by-step explanation:

Let the number of newspaper A sold be 'x'

and the number of newspaper A sold be 'y'

Therefore,

Total revenue = $1x + $1.25y

Total newspapers sold = x + y

Therefore,

according to the question

r = $\frac{x}{x + 1.25y}\times100$ ..........(1)

and,

p = $\frac{x}{x + y}\times100$

or

(x + y)p = 100x

or

y = $\frac{x}{p}\times100-x$

or

y = $\frac{x(100-p)}{p}$

substituting y in 1

r = $\frac{x}{x + 1.25(\frac{x(100-p)}{p})}\times100$

or

r = $\frac{100p}{p+125-1.25p}$

or

r = $\frac{100p}{125-0.25p}$

multiply and divide the RHS with 4

we get

r = (400p)÷(500-p)

hence,

Option 400p/500-p

3 0

3 years ago

1. Find the quartiles for the tire pressures of car tires at an auto clinic. Tire pressure is measured in psi

marusya05 [52]

The quartiles and outliers for the data provided are as follows:

1. Quartiles for the tire pressures of car tires at an auto clinic:

Q2 = 28 psi
Q4 = 36 psi
Q1 = 21.5 psi
Q3 = 34 psi

2. The outlier(s) for the tires prices is $450

3. The outlier(s) for the tires prices is $650

4. The outlier(s) for the tires prices is $540

5. Quartiles for the tire pressures of car tires at an auto clinic are:

Q2 = 27 psi
Q4 = 35 psi
Q1 = 23.5 psi
Q3 = 33 psi

<h3>Quartiles</h3>

Quartiles are the values that divide a list of numbers into four parts or quarters after the numbers are arranged in ascending order.

They describe the dispersionof a set of data values.

Q1 is the first quartile or lower quartile. 25% of the numbers in the data set are at or below Q1.
Q2 is the second quartile. 50% of the numbers are below Q2, and
Q3 is the third quartile, or upper quartile. 75% of the numbers are at or below Q3.
Q4 is the maximum value in the data set. 100% of the numbers are at or below Q4.

<h3>Outliers</h3>

Outliers are data points that differs significantly from other observations.

There are two outliers: lower boundary outliers and upper boundary outliers.

Formula for lower boundary outliers is:

Q1 − 1.5(IQR)

Formula for upper boundary outliers is:

Q3 + 1.5(IQR)

where IQR is interquartile range

IQR = Q3 - Q1

<h3>1. Quartiles for the tire pressures of car tires at an auto clinic: (20, 27,19, 23, 29, 28, 34, 34, 36)</h3>

Arranging in increasing order:

19, 20, 23, 27, 28, 29, 34, 34, 36.

Q2 = 28

Q4 = 36

Q1 = (20 + 23) / 2

Q1 = 21.5

Q3 = (34 + 34) /2

Q3 = 34

<h3>2. Outlier(s) for these tires prices: $58, $95, $78, $125, $87, $158, $152, $182, $195, $450</h3>

Arranging in increasing order:

58, 78, 87, 95, 125, 152, 158, 182, 195, 450

Q1 = (87 + 78)/2

Q1 = 62.5

Q3 = 182 + 195/2

Q3 = 188.5

IQR = 188.5 - 62.5

IQR = 126

Lower boundary outlier(s) = 62.5 - 1.5(126)

Lower boundary outliers = -126.5

Thus there are no lower outliers.

Upper boundary outliers = 188.5 + 1.5(126)

Upper boundary outliers = 377.5

Therefore, 450 is an outlier.

<h3>3. The outlier(s) for these tires prices: $78, $195, $98, $145, $87, $138, $159, $172, $155, $210, $240, $650</h3>

Arranging in increasing order:

78, 87, 98, 138, 145, 155, 159, 172, 195, 210, 240, 650

Q1 = 98

Q3 = 210

IQR = 210 - 98

IQR = 112

Lower boundary outlier(s) = 98 - 1.5(168)

Lower boundary outliers = -70

Thus there are no lower outliers.

Upper boundary outliers = 210 + 1.5(112)

Upper boundary outliers = 378

Therefore, 650 is an outlier.

<h3>4. The outlier(s) for these tires prices: $88, $135, $75, $135, $85, $168, $156, $192, $195, $210, $230, $245, $540</h3>

Arranging in increasing order:

75, 85, 88, 135, 135, 156, 168, 192, 195, 210, 230, 245, 540

Q1 = (88 + 135)/2

Q1 = 111.5

Q3 = (210 + 230)/2

Q3 = 220

IQR = 220 - 111.5

IQR = 108.5

Lower boundary outlier(s) = 111.5 - 1.5(108.5)

Lower boundary outliers = - 51.25

Thus, there are no lower outliers.

Upper boundary outliers = 220 + 1.5(111.5)

Upper boundary outliers = 387.25

Therefore, 540 is an outlier.

<h3>5. Quartiles for the tire pressures of car tires at an auto clinic: 23, 29,18, 24, 27, 24, 35, 32, 34</h3>

Arranging in increasing order:

18, 23, 24, 24, 27, 29, 32, 34, 35.

Q2 = 27

Q4 = 35

Q1 = (23 + 24) / 2

Q1 = 23.5

Q3 = (32 + 34) /2

Q3 = 33

Learn more about quartiles and outliers at: brainly.com/question/24805469

4 0

2 years ago

A bag of Super Whisker cat food, at $12.85, is one third the price of a bag of Power Meow cat food. Write and solve a division e

allsm [11]

Answer:

Price of power meow cat food = $38.55

Step-by-step explanation:

Price of whisker cat food = $12.85

Price of power meow cat food = x

Whisker cat food = ⅓ * power meow cat food

12.85 = ⅓ x

X = 3 * 12.85

X = $38.55

Price of power meow cat food is $38.55

7 0

3 years ago

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