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3 years ago

2hours to 1day as a ratio in its simplest form

Mathematics

2 answers:

Anton [14]3 years ago

8 0

1 day = 24 hours

So,

2 hours : 1 day

2 hours : 24 hours

1 : 12

Murrr4er [49]3 years ago

3 0

Answer:

Step-by-step explanation:

convert 1 day into hours

1 day=24 hour

2/24

=1/12

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Find the equation of the linear function represented by the table y0 5 10 15 x 1 2 3 4

vodomira [7]

Answer: Therefore, your equation of the function will be y = 4x – 4.

Step-by-step explanation:

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2 years ago

The terminal side of an angle, Theta, whose sine value is One-half

Kazeer [188]

Theta has a reference angle of 30° and is in Quadrant I or II.

Sin(theta) = ½

Basic angle: 30

<h3>What is the reference angle?</h3>

The acute angle between the terminal arm/terminal side and the x-axis. The reference angle is always positive. In other words, the reference angle is an angle sandwiched between the terminal side and the x-axis.

Angles: 30,

180-30 = 150

Because sin is positive in quadrants 1 and 2.

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1 year ago

The amounts of nicotine in a certain brand of cigarette are normally distributed with a mean of 0.917 g and a standard deviation

Nutka1998 [239]

Answer:

$z = \frac{0.872-0.917}{\frac{0.303}{\sqrt{37}}}= -0.903$

And we can find this probability to find the answer:

$P(z$

And using the normal standar table or excel we got:

$P(z$

Step-by-step explanation:

Let X the random variable that represent the amounts of nocatine of a population, and for this case we know the distribution for X is given by:

$X \sim N(0.917,0.303)$

Where $\mu=0.917$ and $\sigma=0.303$

We have the following info from a sample of n =37:

$\bar X= 0.872$ the sample mean

And we want to find the following probability:

$P(\bar X \leq 0.872)$

And we can use the z score formula given by;

$z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for the value of 0.872 we got:

$z = \frac{0.872-0.917}{\frac{0.303}{\sqrt{37}}}= -0.903$

And we can find this probability to find the answer:

$P(z$

And using the normal standar table or excel we got:

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6 0

3 years ago

MEC makes a big family tent in the shape of a symmetrical triangular prism. The floor of the tent is 5 m long and 2.5 m wide. Th

SVEN [57.7K]

Answer:

15,236 sq in

Step-by-step explanation:

First, you have to split the net into three rectangles and two triangles

step 1

triangular base area

1/2 bh = 1/2 (44 in)(64 in)

= 1/2 (2,816 sq in)

= 1,408 sq in.

2 x 1,408 sq in = 2,816 sq in.

step 2

end rectangle area

lw = (68 in)(69 in)

= 4,692 sq in

2 x 4,692 sq in = 9,384 sq in

Now find the area of the middle rectangle

lw - (69 in) (44 in) = 3,039 sq in.

then add all the areas together

2,816 + 9,384 + 3,036 = 15,236 sq in

Hope this helps :)

3 0

3 years ago

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

2hours to 1day as a ratio in its simplest form​

2hours to 1day as a ratio in its simplest form