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sp2606 [1]
2 years ago
13

1. Steven purchases a bowl. The diameter of the bowl is 14 cm. What is the circumference of the bowl? 2. Daniel wants to buy coo

kies for her friend. The radius of a cookie is 5 inches. What is the cookie's circumference? 3. Donna makes a round pizza. She wants to put a cheese layer on the pizza. If the cake is 8 cm in diameter, how many square cm of cheese layer does she need to put on the pizza? 4. Cynthia wants to buy a round photo frame for her brother. The radius of the photo frame is 9 cm. What is the photo frame's circumference? 5. Brian made a tasty burger. The diameter of the burger was 5 cm. What was the area of burger?
Mathematics
1 answer:
Pachacha [2.7K]2 years ago
5 0

Answer:

1. The circumference of bowl is 43.98 cm

2. The circumference of cookie is 31.42 inches

3. The amount of square centimetre of cheese layer she needs to put on the pizza is 50.27 cm²

4. The circumference of photo frame is 56.55 cm

5. The area of burger is 19.635 cm²

Step-by-step explanation:

1. Bowl diameter = 14 cm

Circumference of bowl = π·D = π×D

Where:

D = 14 cm

Circumference = π × 14 = 43.98 cm

2. Radius of cookie, r = 5 inches

Circumference of cookie = π·D = π×2×r = π×2×10 = 31.42 inches

3. Diameter, D of pizza = 8 cm

Area of pizza = π·D²/4 = π·8²/4 = 50.27 cm²

The amount of square centimetre of cheese layer she needs to put on the pizza = 50.27 cm²

4. Radius of photo frame = 9 cm

Circumference of photo frame = 2·π·r = 2 × π × 9 = 56.55 cm

5. Diameter, D of burger Brian made = 5 cm

Area of burger is given by the following relation;

Area of burger = π·D²/4 = π·5²/4 = 19.635 cm².

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Answer:

500 calories

Step-by-step explanation:

divide the 1250 calories by the initial five servings and you get 250 then multiply that by the 2 servings they are asking for and you get 500 calories

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2 years ago
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A product is composed of four parts. In order for the product to function properly in a given situation, each of the parts must
svet-max [94.6K]

Answer:

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Step-by-step explanation:

For the product to work, all four probabilities must come to pass, so that

P(Part-1)*P(Part-2)*P(Part-3)*P(Part-4)

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P(Part-1) = 0.96

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As all parts are independent, so the formula is P(A∩B) = P(A)*P(B)

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P (Working Product) = 0.96*0.96*0.96*0.99*0.99

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2 years ago
What percent of 2b is: 0.04b; 0.2b; 0.56b; 1.8b; 2.5b; 3b; By what percent are they each larger or smaller than 2b
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.04b is 2% of 2b because .04 multiplied by 100 is 4 then you divide by 2 which gets you 2.
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it is 98% smaller than 2b.

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2 years ago
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Mrs.Steffen’s third grade class has 30 students in it. The students are divided into three groups(numbered 1, 2,and 3),each havin
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Answer:

a. \\ 10! = 3628800;

b. \\ 10!*10!*10! = 47784725839872000000 = 4.7784725839872*10^{19}

Step-by-step explanation:

We need here to apply the <em>Multiplication Principle </em>or the <em>Fundamental Principle of Counting</em> for each answer. Answer <em>b</em> needs an extra reasoning for being completed.

The <em>Multiplication Principle</em> states that if there are <em>n</em> ways of doing something and <em>m</em> ways of doing another thing, then there are <em>n</em> x <em>m</em> ways of doing both (<em>Rule of product</em> (2020), in Wikipedia).

<h3>In how many ways can ten students line up? </h3>

There are <em>ten</em> students. When one is selected, there is no other way to select it again. So, <em>no repetition</em> is allowed.

Then, in the beginning, there are 10 possibilities for 10 students; when one is selected, there are nine possibilities left. When another is selected, eight possibilities are left to form the file, and so on.

Thus, we need to multiply the possibilities after each selection: that is <em>why</em> the <em>Multiplication Principle</em> is important here.

This could be expressed mathematically using n!:

\\ n! = n * (n-1)! * (n-2)! *...* 2*1.

For instance, \\ 5! = 5 * (5-1)! * (5-2)! *...*2*1 = 5 * 4 * 3 * 2 * 1 = 120.

So, for the case in question, the <em>ten</em> students can line up in:

\\ 10! = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 3628800 ways to line up in a single file.

<h3>Second Question</h3>

For this question, we need to consider the former reasoning with extra consideration in mind.

The members of Group 1 can occupy <em>only</em> the following places in forming the file:

\\ G1 = \{ 1, 4, 7, 10, 13, 16, 19, 22, 25, 28\}^{th} <em>places</em>.

The members of Group 2 <em>only</em>:

\\ G2 = \{ 2, 5, 8, 11, 14, 17, 20, 23, 26, 29\}^{th} <em>places</em>.

And the members of Group 3, the following <em>only</em> ones:

\\ G3 = \{ 3, 6, 9, 12, 15, 18, 21, 24, 27, 30\}^{th} <em>places.</em>

Well, having into account these possible places for each member of G1, G2 and G3, there are: <em>10! ways</em> for lining up members of G1; <em>10! ways</em> for lining up members of G2 and, also, <em>10! ways</em> for lining up members of G3.

After using the <em>Multiplication Principle</em>, we have, thus:

\\ 10! * 10! * 10! = 47784725839872000000 = 4.7784725839872 *10^{19} <em>ways the students can line up to come in from recess</em>.

3 0
1 year ago
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2 years ago
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