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2 years ago

a spinner is divided into 10 equal sections numbered from 1 to 10. You have spin the spinner once. What is P (divisible by 2)

1 answer:

6 0

There is a 1/2 probability that the number will be divisible by 2 because there are 5 numbers from 1 to 10 divisible by 2. So 5/10 = 1/2.

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8^2 x (2+6) / 4 solve

Softa [21]

Answer:

$8^2\times \frac{\left(2+6\right)}{4}$

$\frac{\left(2+6\right)}{4}$ = 8/4 = 2

$8^2 \times 2$ =

64 x 2 =

128.

3 0

2 years ago

Please check my question. 11 points. thanks for the help.

n200080 [17]

Its right.................

7 0

3 years ago

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

Marrrta [24]

Split up the interval [0, 2] into n 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 n 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

2 years ago

Hey! pls help i’ll give brainliest

grandymaker [24]

Answer:

landslide

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Find the part of the number. * Find 3 1/16 of 32.

oee [108]

Answer:

Step-by-step explanation:

6 0

3 years ago