How do I simplify this expression?

Use inverse operations to complete the second equation each time

ratelena [41]

Answer:

a) 55-42=13

b) 29-10=19

c) 65÷5=13

d) 72÷8=9

6 0

2 years ago

If you have 66 data observations in a sample, how many classes (bins) does the sturges' rule recommend for you to construct a fr

Iteru [2.4K]

According to Sturge's rule, number of classes or bins recommended to construct a frequency distribution is k ≈ 7

Sturge's Rule: There are no hard and fast guidelines for the size of a class interval or bin when building a frequency distribution table. However, Sturge's rule offers advice on how many intervals one can make if one is genuinely unable to choose a class width. Sturge's rule advises that the class interval number be for a set of n observations.

Given,

n = 66

We know that,

According to Sturge's rule, the optimal number of class intervals can be determined by using the equation:

$k=1+log_{2} n$

Here, n is equal to 66 and by substituting the value to the equation we get:

$k=1+log_{2} (66)$

k = 7.0444

k ≈ 7

Learn more about Sturge's rule here: brainly.com/question/28184369

#SPJ4

8 0

10 months ago

The sum of the measures of angle LMN and angle NMP is 180 degrees

Vaselesa [24]

Answer:

angle LMN is 153 degrees.

Step-by-step explanation:

180 = 15g + 18 +3g; combine like terms

180 = 18g + 18; subtract 18 on both sides

162 = 18g; divide 18 on both sides

g = 9

plug in: 15(9) + 18 = 153

please mark brainliest if I helped you!

6 0

2 years ago

I'm having trouble with #2. I've got it down to the part where it would be the integral of 5cos^3(pheta)/sin(pheta). I'm not sur

Butoxors [25]

$\displaystyle\int\frac{\sqrt{25-x^2}}x\,\mathrm dx$

Setting $x=5\sin\theta$ , you have $\mathrm dx=5\cos\theta\,\mathrm d\theta$ . Then the integral becomes

$\displaystyle\int\frac{\sqrt{25-(5\sin\theta)^2}}{5\sin\theta}5\cos\theta\,\mathrm d\theta$
$\displaystyle\int\sqrt{25-25\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\sqrt{1-\sin^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$

Now, $\sqrt{x^2}=|x|$ in general. But since we want our substitution $x=5\sin\theta$ to be invertible, we are tacitly assuming that we're working over a restricted domain. In particular, this means $\theta=\sin^{-1}\dfrac x5$ , which implies that $\left|\dfrac x5\right|\le1$ , or equivalently that $|\theta|\le\dfrac\pi2$ . Over this domain, $\cos\theta\ge0$ , so $\sqrt{\cos^2\theta}=|\cos\theta|=\cos\theta$ .

Long story short, this allows us to go from

$\displaystyle5\int\sqrt{\cos^2\theta}\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$

to

$\displaystyle5\int\cos\theta\dfrac{\cos\theta}{\sin\theta}\,\mathrm d\theta$
$\displaystyle5\int\dfrac{\cos^2\theta}{\sin\theta}\,\mathrm d\theta$

Computing the remaining integral isn't difficult. Expand the numerator with the Pythagorean identity to get

$\dfrac{\cos^2\theta}{\sin\theta}=\dfrac{1-\sin^2\theta}{\sin\theta}=\csc\theta-\sin\theta$

Then integrate term-by-term to get

$\displaystyle5\left(\int\csc\theta\,\mathrm d\theta-\int\sin\theta\,\mathrm d\theta\right)$
$=-5\ln|\csc\theta+\cot\theta|+\cos\theta+C$

Now undo the substitution to get the antiderivative back in terms of $x$ .

$=-5\ln\left|\csc\left(\sin^{-1}\dfrac x5\right)+\cot\left(\sin^{-1}\dfrac x5\right)\right|+\cos\left(\sin^{-1}\dfrac x5\right)+C$

and using basic trigonometric properties (e.g. Pythagorean theorem) this reduces to

$=-5\ln\left|\dfrac{5+\sqrt{25-x^2}}x\right|+\sqrt{25-x^2}+C$

4 0

2 years ago

Read 2 more answers

Which of the following are not trigonometry identities? Check all that apply

Anni [7]

B and A

That’s the answer bro trust xx

6 0

1 year ago