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3 years ago

There are 399 items advertised in a catalog. If the catalog is 75 pages long, what is the average number of items per page?

Mathematics

1 answer:

sukhopar [10]3 years ago

5 0

Answer:

Step-by-step explanation:

Well, we can divide 399 and 75 so it would be 5.32. Round that and it would be about 5 items per page

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Solve the compound inequality: 3x ≤ 18 AND x + 4 > 2

dedylja [7]

There must be a mistake in your question:
If 3 x ≤ 9, then : x ≤ 3 and :
x + 4 > 2
x > 2 - 4
x > - 2
Answer: D)
Solution set : - 2 < x ≤ 3

4 0

3 years ago

Read 2 more answers

What is interest earned on 3500 invested at 6% compounded quarterly or 12 years

enyata [817]

$\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$3500\\
r=rate\to 6\%\to \frac{6}{100}\to &0.06\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{quarterly, thus four}
\end{array}\to &4\\
t=years\to &12
\end{cases}
\\\\\\
A=3500\left(1+\frac{0.06}{4}\right)^{4\cdot 12}\implies A=3500(1.015)^{48}$

so, the interest earned then will be A - 3500.

6 0

3 years ago

Four runners start at the same point; Lin, Elena, Diego, Andre. For each runner write a multiplication equation that describes t

Burka [1]

Answer:

Lin: 25 × 8,2 = 205 meters

Elena: 250 ÷ 28 = 8,9 meters per second

Diego: 32 × (-8,1) = -250,2 meters

Andre: -258 ÷ 25 = -10,32 meters per second

7 0

2 years ago

A moose population is growing exponentially following the pattern in the table shown below. Assuming that the pattern continues,

Alika [10]

The population of the moose after 12 years is 7692

Step-by-step explanation:

The form of the exponential growth function is $y=a(b)^{x}$ , where

a is the initial value
b is the growth factor

The table:

→ x : y

→ 0 : 40

→ 1 : 62

→ 2 : 96

→ 3 : 149

→ 4 : 231

To find the value of a , b in the equation use the data in the table

∵ At x = 0 , y = 40

- Substitute them in the form of the equation above

∵ $40=a(b)^{0}$

- Remember any number to the power of zero is 1 (except 0)

∴ 40 = a(1)

∴ a = 40

- Substitute the value of a in the equation

∴ $y=40(b)^{x}$

∵ At x = 1 , y = 62

- Substitute them in the form of the equation above

∵ $62=40(b)^{1}$

∴ 62 = 40 b

- Divide both sides by 40

∴ 1.55 ≅ b

∴ The growth factor is about 1.55

- Substitute its value in the equation

∴ The equation of the population is $y=40(1.55)^{x}$

To find the population of moose after 12 years substitute x by 12 in the equation

∵ $y=40(1.55)^{12}$

∴ y ≅ 7692

The population of the moose after 12 years is 7692

Learn more:

You can learn more about the logarithmic functions in brainly.com/question/11921476

#LearnwithBrainly

7 0

3 years ago

Write 5 to the power of 8 as a quotient of two exponential terms with the same base in four different ways. use negative or zero

ryzh [129]

So, we are given 5^8. It was happy and content. But then... we had to write it as a quotient of two exponential terms with the same base in four different ways and use negative or zero exponents and ahhhhhh!!!

... anyways...

We'll build a quotient of two exponential terms with the same base 5. Something like this:

5^a / 5^b

We need them to make 5^8 when we are done. I'll first use a zero exponent.

[1] Now, zero exponents are nice since they make things equal 1. Like 5^0 = 1. Well, obviously, 5^8 / 1 = 5^8. So, our first quotient can be:

5^8 / 5^0

Done.

[2] Let's try this on its head. This one's a little weird. Remember that negative exponents flip things upside down. So 5^-8 = 1/5^8 and 1/5^-8 = 5^8 for example. In fact... that's the answer!

5^0 / 5^-8 = 5^8

Done.

[3] Let's try to not use 0s or 8s. We can be clever and do something like this:

5^-1 / 5^-9

What the heck is that? Well, we just flip them and get:

5^-1 / 5^-9 = 5^9 / 5^1 = 5^8

Done.

[4] Can you come up with one last trick on your own? Try it!

3 0

4 years ago