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3 years ago

Gianna is converting a 12-foot-by-15-foot room in her

Mathematics

1 answer:

givi [52]3 years ago

5 0

Answer:

is there a picture of the question

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Who created the circumference of the circle

densk [106]

Abraham Lincoln djdjxu d d d d d d djdjsisow s susbs s s

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4 years ago

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Of all the Sunny Club members in a particular city, 25% prefer swimming on weekends and 75% prefer swimming on weekdays. It is f

Ivan

P(f | weekend) = p(f & weekend)/p(weekend)
.. = 10%/25%
.. = 2/5 = 0.4

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3 years ago

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How do you answer this pls help 7^2+4^3÷2^5

lana66690 [7]

Assignment: $\bold{Solve \ Equation: \ 7^2 \ + \ 4^3 \ \div 2^5}$

<><><><><><><>

Answer: $\boxed{\bold{51}}$

<><><><><><><>

Explanation: $\downarrow\downarrow\downarrow$

<><><><><><><>

[ Step One ] Apply PEMDAS Order Of Operations

Note: $\bold{PEMDAS: \ Parenthesis, \ Exponents, \ Multiply, \ Divide, \ Add, \ Subtract}$

$\bold{7^2: \ 49}$

[ Step Two ] Rewrite Equation

$\bold{49+64\div \:2^5}$

[ Step Three ] Calculate Exponents

$\bold{4^3: \ 64}$

[ Step Four ] Rewrite Equation

$\bold{49+64\div \:2^5}$

[ Step Five ] Calculate Exponents

$\bold{2^5: \ 32}$

[ Step Six ] Rewrite Equation

$\bold{49+64\div \:32}$

[ Step Seven ] Divide

$\bold{64\div \:32: \ 2}$

[ Step Eight ] Rewrite Equation

$\bold{49+2}$

[ Step Nine ] Add

$\bold{51}$

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$\bold{\rightarrow Mordancy \leftarrow}$

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4 years ago

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A class contains 3 female and 9 male students. Find the probability that a student chosen at random is a male, and then a second

sammy [17]

9 +3 = 12 students

9 are male
1st = 9/12 = 3/4
2nd = 8/11

3/4 * 8/11 = 6/11

Answer = D) 6/11

8 0

3 years ago

Serhud [2]

A circle is characterized by radius, arc, sectors and circumference

The length of the major arc is $18.75\pi$
The radius of the circle is 15
The area of the shaded sector is $140.625 * \pi$

<h3>Length of the major arc</h3>

The given parameters are:

$C = 30\pi$ --- the circumference

$\theta = 225^o$ -- the center angle

The length of the major arc is calculated using:

$L = \frac{\theta}{360} * C$

So, we have:

$L = \frac{225}{360} *30\pi$

Evaluate

$L = 18.75\pi$

Hence, the length of the major arc is $18.75\pi$

<h3>The radius of the circle</h3>

The circumference is given as:

$C = 30\pi$

So, we have:

$2\pi r = 30\pi$

Divide through by 2pi

$r = 15$

Hence, the radius of the circle is 15

<h3>The area of the shaded sector</h3>

The area of a sector is:

$A = \frac{\theta}{360} * \pi r^2$

So, we have:

$A = \frac{225}{360} * \pi * 15^2$

Evaluate

$A = 140.625 * \pi$

Hence, the area of the shaded sector is $140.625 * \pi$