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3 years ago

It takes 3.5 ounces of cocoa

Mathematics

1 answer:

schepotkina [342]3 years ago

5 0

Answer:

24.5 oz

Step-by-step explanation:

1 cup ----> required 3.5 oz of cocoa

1 cups -----> requires 3.5 oz x 7 = 24.5 oz of cocoa

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41,42,43,44,45,45,45,45 what number is the median? what number is the mode?

Assoli18 [71]

Answer: Median: 44.5 Mode: 45

Step-by-step explanation: Median is the middle number. Since it's an even set, you add the two middle numbers and divide by two. The mode is just the number that appears the most.

4 0

2 years ago

wolverine [178]

Answer:

Good Luck!

Step-by-step explanation:

6 0

2 years ago

“An architect is planning to incorporate several stone spheres of different sizes into the landscaping of a public park, and wor

Artist 52 [7]

Answer:

Part 1) The value that is closest to the cost of finishing a sphere with a 5.50-meter circumference is $900

Part 2) The value that is closest to the cost of finishing a sphere with a 7.85-meter circumference is $1,800

Step-by-step explanation:

Step 1

Find the radius of each sphere

we know that

The circumference of a circle is equal to

$C=2\pi r$

Find the radius of the sphere with a 5.50-meter circumference

For $C=5.50\ m$

assume

$\pi =3.14$

substitute and solve for r

$5.50=2(3.14)r$

$r=5.50/[2(3.14)]=0.88\ m$

Find the radius of the sphere with a 7.85-meter circumference

For $C=7.85\ m$

assume

$\pi =3.14$

substitute and solve for r

$7.85=2(3.14)r$

$r=7.85/[2(3.14)]=1.25\ m$

step 2

Find the surface area of each sphere

The surface area of sphere is equal to

$SA=4\pi r^{2}$

Find the surface area of sphere with a 5.50-meter circumference

For $r=0.88\ m$

assume

$\pi =3.14$

substitute

$SA=4(3.14)(0.88)^{2}$

$SA=9.73\ m^{2}$

Find the surface area of sphere with a 7.85-meter circumference

For $r=1.25\ m$

assume

$\pi =3.14$

substitute

$SA=4(3.14)(1.25)^{2}$

$SA=19.63\ m^{2}$

step 3

Find the cost of finishing each sphere

we know that

To find out the cost , multiply the surface area by $92 per square meter

Find the cost of sphere with a 5.50-meter circumference

$9.73*(92)=\$895.16$

therefore

The value that is closest to the cost of finishing a sphere with a 5.50-meter circumference is $900

Find the cost of sphere with a 7.85-meter circumference

$19.63*(92)=\$1,805.96$

therefore

The value that is closest to the cost of finishing a sphere with a 7.85-meter circumference is $1,800

6 0

3 years ago

The change from one number to the next in a geometric sequence is called

Dmitrij [34]

Answer:

A) or B)

Step-by-step explanation:

Im not really sure but its one of them

7 0

3 years ago

2. The ratio of

denpristay [2]

Answer:

Step-by-step explanation:

Given the ratio of the profit, cost of materials and labour in the production of

an article to be 5:7:13 respectively, total ratio = 5+7+13 = 25

If the cost of labour is x, the cost of material will be 840+x (since the cost of materials is Le 840 more than that of labour) .

Let the total cost of producing the article be y.

Cost of labour = 13/25 * y = x

Cost of labour = 13y/25 = x.................... 1

Cost of material = 7/25*y = 840+x

Cost of material = 7y/25 = 840+x ..................... 2

From 1, 13y = 25x

x = 13y/25 ................... 3

Substituting equation 3 into 2:

7y/25 = 840+x

7y/25 = 840+13y/25

collect the like terms:

7y/25 - 13y/25 = 840

-6y/25 = 840

-6y = 25*840

y = 25*840/6

y = 3,500

Hence the total cost of producing the article is 3500

4 0

3 years ago