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3 years ago

Which figure shows 25% shaded?

Mathematics

1 answer:

sleet_krkn [62]3 years ago

6 0

25% is like cutting a piece of pie or pizza into 4 equal slices. One of those slices is 25% Because 25% is 1/4

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if 1/2 gallons of paint covers 1/4 of a wall then how much paint will be needed to cover the entire wall

belka [17]

2 gallons of paint are required!

Hope this helps!!

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3 years ago

Clarise evaluated this expression. (66.3 – 14.62) ÷ 0.6 – 0.22 (51.68) ÷ 0.6 – 0.22 (51.68) ÷ 0.42 51.68 ÷ 0.16 32.3 Which error

Pepsi [2]

The error she make is the last one 0.1632.3

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4 years ago

Read 2 more answers

There's a room, the height is 12, length is 14, and width is 10. What is the total area of the 4 walls?

Scorpion4ik [409]

Answer:

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Step-by-step explanation:

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3 years ago

Diana put $8000 in a 10-year CDpaying 5% interest compoundedmonthly. After 2 years, shewithdrew all her money, and as anearly wi

Svet_ta [14]

Answer:

$8,430.23

Explanation:

From the statement of the problem:

• The principal amount = $8,000

• Interest Rate = 5%

• Compounding Period = 12 (Monthly)

The compound interest formula is given as:

$A(n)=P\left(1+\frac{r}{k}\right)^{nk}$

Using the compound period formula, we first, calculate the amount in her account at the end of 1 year.

$\begin{gathered} A(1)=8000\left(1+\frac{0.05}{12}\right)^{12\times1} \\ A(1)=\$8409.30 \end{gathered}$

This means that the interest she made during the first year is:

$\text{ Interest during the first year}=8409.30-8000=\$409.30$

Next, calculate the amount in her account at the end of the second year.

$\begin{gathered} A(2)=8000\left(1+\frac{0.05}{12}\right)^{12\times2} \\ A(2)=\$8839.53 \end{gathered}$

Since she paid back all the interest she made during the first year, the amount Diana was left with is:

$\begin{gathered} 8839.53-409.30 \\ =8,430.23 \end{gathered}$

Diana was left with $8,430.23.

6 0

1 year ago

A tennis tournament has 2n contestants. We want to pair them up for the first round of singles matches. Show that the number of

ddd [48]

There are

$\dbinom{2n}2 = \dfrac{(2n)!}{2! (2n-2)!}$

ways of pairing up any 2 members from the pool of $2n$ contestants. Note that

$(2n)! = 1\times2\times3\times4\times\cdots\times(2n-2)\times(2n-1)\times(2n) = (2n-2)! \times(2n-1) \times(2n)$

so that

$\dbinom{2n}2 = \dfrac{(2n)\times(2n-1)\times(2n-2)!}{2! (2n-2)!} = \boxed{n(2n-1)}$

4 0

2 years ago