All categories

3 years ago

What is greater 2 3/4 or 2 2/3

Mathematics

2 answers:

Ronch [10]3 years ago

5 0

2 3/4 is greater then 2 2/3

yKpoI14uk [10]3 years ago

4 0

2 * 3/4 = 1.5

2 * 2/3 = 1 1/3

1.5 > 1 1/3

2 3/4 > 2 2/3

You might be interested in

In my number the digit in the ones place is double the digit in the tens place. the sum of the digits is 3

jasenka [17]

The number is 21 because 2 is 1 doubled, and 2+1 is 3.

6 0

2 years ago

We draw a card from a standard deck. What is the probability that it is a Jack, Queen, King or it is red.

enyata [817]

Answer:

8/13 chance to get a jack, Queen, King or Red

3 0

3 years ago

The price of two pairs and six apples is $14. The price of three pairs and nine apples is $21. Can you determine the unit prices

deff fn [24]

Answer:

Step-by-step explanation:

The second purchase is exactly half-again larger than the first purchase, so sheds no light on the relative costs of the items. The per-item costs can be found when the ratio of items purchased is different from one buy to another.

___

We can only say that the cost of 1 pear and 3 apples is $7.

6 0

3 years ago

How many solutions does this equation have?

Leto [7]

Answer:

The given equation having only one solution i.e., "w = 3". A further explanation is provided below.

Step-by-step explanation:

The given equation is:

⇒ $9(w-9)-2=-7w+7(w-8)$

By opening the brackets, we get

⇒ $9w-81-2=-7w+7w-56$

⇒ $9w-83=-56$

On adding "83" both sides, we get

⇒ $9w-83+83=-56+83$

⇒ $9w=27$

⇒ $w=\frac{27}{9}$

⇒ $=3$

4 0

2 years ago

You are saving money to buy an electric guitar. You deposit $1000 in an account that earns interest compounded annually. The exp

Katena32 [7]

Let's move like a crab, backwards some.

after 2 years?

$\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$1000\\
r=rate\to 3\%\to \frac{3}{100}\to &0.03\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &2
\end{cases}
\\\\\\
A=1000\left(1+\frac{0.03}{1}\right)^{1\cdot 2}\implies A=1000(1.03)^2$

after 3 years?

$\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$1000\\
r=rate\to 3\%\to \frac{3}{100}\to &0.03\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &3
\end{cases}
\\\\\\
A=1000\left(1+\frac{0.03}{1}\right)^{1\cdot 3}\implies A=1000(1.03)^3$

is that enough to pay the $1100?

now, let's write 1000(1+r)² in standard form

1000( 1² + 2r + r²)

1000(1 + 2r + r²)

1000 + 2000r + 1000r²

1000r² + 2000r + 1000 <---- standard form.

8 0

3 years ago