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podryga [215]
3 years ago
15

Multiple choice please answer please

Mathematics
1 answer:
tankabanditka [31]3 years ago
3 0

Answer:

B.cos^{-1} \bigg(\frac{4}{5}\bigg)

Step-by-step explanation:

\because \cos \angle B = \bigg(\frac{4}{5}\bigg)\\\\\therefore  \angle B =cos^{-1} \bigg(\frac{4}{5}\bigg)\\

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Name the transformations of the next equation y=3(x+6) - 5
klemol [59]

Answer:

vertical stretch

Step-by-step explanation:

8 0
2 years ago
mt. Everest, the highest elevation is Asia is 29028 feet above see level. the dead sea, the lowest elevation is 1312 feet below
marshall27 [118]
29028 + 1312 = 30340
5 0
3 years ago
Given that it was less than 80º on a given day, what is the probability that it also rained that day? 0.3 0.35 0.65 0.7
eimsori [14]

The probability that it also rained that day is to be considered as the 0.30 and the same is to be considered.

<h3>What is probability?</h3>

The extent to which an event is likely to occur, measured by the ratio of the favorable cases to the whole number of cases possible.

The probability that the temperature is lower than 80°F and it rained can be measured by determining the number at the intersection of a temperature that less than 80°F and rain.

So, This number is 0.30.

Hence, we can say that it was less than 80°F on a given day, the probability that it also rained that day is 0.30.

To learn more about the probability from the given link:

brainly.com/question/18638636

The above question is incomplete.

The conditional relative frequency table was generated using data that compared the outside temperature each day to whether it rained that day. A 4-column table with 3 rows titled weather. The first column has no label with entries 80 degrees F, less than 80 degrees F, total. The second column is labeled rain with entries 0.35, 0.3, nearly equal to 0.33. The third column is labeled no rain with entries 0.65, 0.7, nearly equal to 0.67. The fourth column is labeled total with entries 1.0, 1.0, 1.0. Given that it was less than 80 degrees F on a given day, what is the probability that it also rained that day?

#SPJ4

7 0
2 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
What are the first ten multiples of 12
iris [78.8K]
12x1
12x2
12x3
12x4
12x5
12x6
12x7
12x8
12x9
12x10
4 0
3 years ago
Read 2 more answers
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