What fraction is bigger 1/4 or 16 1/3

Solve each equation mentally.

larisa [96]

Answer:

1) 1/17

2) 1/111

3) 1/15

They're all fractions

Hope that helped, just divide the result with the number available

Example :

1÷17 =1/17

4 0

3 years ago

Carl earns $8 per hour in his summer job. He hopes to earn between $120 and $200 per week. How many hours (x) will he need to wo

yanalaym [24]

He would need to work

120/8=15
orr
120x8=960

8 0

3 years ago

Read 2 more answers

A college library has five copies of a certain text on reserve. Three copies (1, 2, and 3) are first printings, and the remainin

dezoksy [38]

Answer:

S = { (4), (5), (1,4), (1,5), (2,4), (2,5), (3,4), (3,5), (1,2,3,4), (1,2,3,5), (2,1,3,4) (2,1,3,5), (3,1,2,4), (3,1,2,5) }

A ={(4), (5)}

Step-by-step explanation:

Given that:

Among the three copies, (1,2,3) are the first printings, and (4,5) are the second printings.

A student who examines these books in random order stops when a second printing has been selected.

Thus, we can compute the sample space associated with these experiments as:

S = { (4), (5), (1,4), (1,5), (2,4), (2,5), (3,4), (3,5), (1,2,3,4), (1,2,3,5), (2,1,3,4) (2,1,3,5), (3,1,2,4), (3,1,2,5) }

Suppose A represents the event that we must examine exactly one book.

Then the outcomes of A are:

A ={(4), (5)}

6 0

3 years ago

You spin the spinner once.

Ymorist [56]

Answer:

0.625

Step-by-step explanation:

5 out of 8 of the sections in the circle are yellow. This means that there is a 5/8 chance of the spinner landing in a yellow section.

5/8 written as a decimal is 0.625

The answer is 0.625

6 0

3 years ago

PLZ HELP!!! Use limits to evaluate the integral.

Marrrta [24]

Split up the interval [0, 2] into <em>n</em> equally spaced subintervals:

$\left[0,\dfrac2n\right],\left[\dfrac2n,\dfrac4n\right],\left[\dfrac4n,\dfrac6n\right],\ldots,\left[\dfrac{2(n-1)}n,2\right]$

Let's use the right endpoints as our sampling points; they are given by the arithmetic sequence,

$r_i=\dfrac{2i}n$

where $1\le i\le n$ . Each interval has length $\Delta x_i=\frac{2-0}n=\frac2n$ .

At these sampling points, the function takes on values of

$f(r_i)=7{r_i}^3=7\left(\dfrac{2i}n\right)^3=\dfrac{56i^3}{n^3}$

We approximate the integral with the Riemann sum:

$\displaystyle\sum_{i=1}^nf(r_i)\Delta x_i=\frac{112}n\sum_{i=1}^ni^3$

Recall that

$\displaystyle\sum_{i=1}^ni^3=\frac{n^2(n+1)^2}4$

so that the sum reduces to

$\displaystyle\sum_{i=1}^nf(r_i)\Delta x_i=\frac{28n^2(n+1)^2}{n^4}$

Take the limit as <em>n</em> approaches infinity, and the Riemann sum converges to the value of the integral:

$\displaystyle\int_0^27x^3\,\mathrm dx=\lim_{n\to\infty}\frac{28n^2(n+1)^2}{n^4}=\boxed{28}$

Just to check:

$\displaystyle\int_0^27x^3\,\mathrm dx=\frac{7x^4}4\bigg|_0^2=\frac{7\cdot2^4}4=28$

4 0

3 years ago