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3 years ago

Covert 50% into a decimal

Mathematics

1 answer:

BigorU [14]3 years ago

3 0

Converting From Percent to Decimal

When we divide 50 by 100 we get 0.5 (a decimal number).

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What is the missing number in this pattern?<br> 0, 2, 6, ----, 30, 62

mario62 [17]

Answer:

Step-by-step explanation:

so the pattern is they add 2 then they add 4 then they add 8 then the add 16 basically they are double the number being added each number so the missing number is 14

5 0

2 years ago

Read 2 more answers

if every child took 12 apples,there would be three left.if each childtook 9 apples,there would be 21 apples left how many apples

White raven [17]

Multiples of 12 : 12 , 24 , 36 , 42 , 60 , 72 , 84 , 96 , 108 , 120
3 left (+3) : 15 , 27 , 39 , 45 , 63 , 81 , 87 , 99 , 111 , 123

Multiples of 9 : 09 , 18 , 27 , 36 , 45 , 54 , 64 , 72 , 81 , 90
21 left (+21) : 30 , 39 , 48 , 57 , 66 , 75 , 85 , 93 , 102, 111

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Answer : There are 111 apples
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5 0

3 years ago

In a probability experiment, Eric flipped a coin 39 times. The coin landed on heads 26 times. What is the ratio of heads to tail

PilotLPTM [1.2K]

Answer:

26 : 13

Explanation:

Total # of times the coin was flipped: 39

# of heads: 26

39 - 26 = 13

# of tails: 13

Heads : Tails

26 : 13

4 0

3 years ago

Need help it seems so simple but i dont know what it is

wel

3,5 ?
Hope this helps!

4 0

3 years ago

Read 2 more answers

Use the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule to approximate the given integral with the specified value of n.

Vera_Pavlovna [14]

Split up the integration interval into 4 subintervals:

$\left[0,\dfrac\pi8\right],\left[\dfrac\pi8,\dfrac\pi4\right],\left[\dfrac\pi4,\dfrac{3\pi}8\right],\left[\dfrac{3\pi}8,\dfrac\pi2\right]$

The left and right endpoints of the $i$ -th subinterval, respectively, are

$\ell_i=\dfrac{i-1}4\left(\dfrac\pi2-0\right)=\dfrac{(i-1)\pi}8$

$r_i=\dfrac i4\left(\dfrac\pi2-0\right)=\dfrac{i\pi}8$

for $1\le i\le4$ , and the respective midpoints are

$m_i=\dfrac{\ell_i+r_i}2=\dfrac{(2i-1)\pi}8$

Trapezoidal rule

We approximate the (signed) area under the curve over each subinterval by

$T_i=\dfrac{f(\ell_i)+f(r_i)}2(\ell_i-r_i)$

so that

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4T_i\approx\boxed{3.038078}$

Midpoint rule

We approximate the area for each subinterval by

$M_i=f(m_i)(\ell_i-r_i)$

so that

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4M_i\approx\boxed{2.981137}$

Simpson's rule

We first interpolate the integrand over each subinterval by a quadratic polynomial $p_i(x)$ , where

$p_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m)\dfrac{(x-\ell_i)(x-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(x-\ell_i)(x-m_i)}{(r_i-\ell_i)(r_i-m_i)}$

so that

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx$

It so happens that the integral of $p_i(x)$ reduces nicely to the form you're probably more familiar with,

$S_i=\displaystyle\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx=\frac{r_i-\ell_i}6(f(\ell_i)+4f(m_i)+f(r_i))$

Then the integral is approximately

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4S_i\approx\boxed{3.000117}$

Compare these to the actual value of the integral, 3. I've included plots of the approximations below.

3 0

3 years ago