All categories

3 years ago

I NEED THIS ANSWERED ASAP PLEASE HELP!!! select the correct answer for points

Mathematics

1 answer:

krek1111 [17]3 years ago

5 0

Answer:

C. 120°

Step-by-step explanation:

length of arc = 4π cm

r = 6 cm

central angle = s/r

=> central angle = 4π/6

=> central angle = 2π/3 radians

In degrees,

central angle = (2π/3)(180/π) = 120°

hope this helps and is right. p.s i really need brainliest :)

You might be interested in

I’m literally begging someone please help (just number 10. show work)

Sladkaya [172]

Answer:

$h = \frac{2A}{b +B }$

Height= 16 feet

Step-by-step explanation:

Please see attached picture for full solution.

6 0

3 years ago

Who wants brainllest follow my tik tok .splited

sergey [27]

Answer: do you do f4f?

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

With double-digit annual percentage increases in the cost of health insurance, more and more workers are likely to lack health i

Dahasolnce [82]

Answer:

1. χ² = 15.3902

2. The p value is :_____.a. less than .005

3. Conclusion

a. health insurance coverage is not independent of the size of the company

4. The percentages of employees

Small %= 33/50= 0.66

Medium %= 68/75 = 0.91

Large %= 88/100= 0.88

Step-by-step explanation:

1) We set up our null and alternative hypothesis as

H0: the employee health insurance coverage is independent of the size of the company

against the claim

Ha: the employee health insurance coverage is not independent of the size of the company

2) the significance level alpha is set at 0.05

3) the test statistic under H0 is

χ²= ∑ (O - E)²/ E where O is the observed and E is the expected frequency

which has an approximate chi square distribution with ( 3-1) (2-1)= 2 d.f

4) Computations:

Under H0 , the observed frequencies are :

Observed Expected E (O-E) (O-E)² (O-E)²/E

33 42 -9 81 1.928

68 63 5 25 0.3968

88 84 4 16 0.1904

17 8 9 81 10.125

7 12 -5 21 1.75

12 16 -4 16 1

15.3902

Expected Values are calculated using the formula :

Row Total * Column Total / sample size

E1= (33+17) (33+ 68+88)/ 50+75+100= 42

E4= (33+17) (17+ 7+ 12)/ 50+75+100=8

E2= (68+7) (33+ 68+88)/ 50+75+100= 63

E5= (68+7) (17+ 7+ 12)/ 50+75+100= 12

E3= (88+12) (33+ 68+88)/ 50+75+100= 84

E6= (88+12) (17+ 7+ 12)/ 50+75+100= 16

5) The critical region is χ² ≥ χ² (0.05)2 = 5.99

6) Conclusion:

The calculated χ² = 15.3902 falls in the critical region χ² ≥ 5.99 so we reject the null hypothesis that the employee health insurance coverage is NOT independent of the size of the company.

2. The p value is :_____.

a. less than .005

The p-value is .000385.

3. Conclusion

a. health insurance coverage is not independent of the size of the company

4. The percentages of employees

Small %= 33/50= 0.66

Medium %= 68/75 = 0.91

Large %= 88/100= 0.88

3 0

3 years ago

Factor 8x^5+4x^2−12

Effectus [21]

D) because only 4 is the common factor in the equation.

4 0

3 years ago

Read 2 more answers

Let X ~ N(0, 1) and Y = eX. Y is called a log-normal random variable.

Cloud [144]

If $F_Y(y)$ is the cumulative distribution function for $Y$ , then

$F_Y(y)=P(Y\le y)=P(e^X\le y)=P(X\le\ln y)=F_X(\ln y)$

Then the probability density function for $Y$ is $f_Y(y)={F_Y}'(y)$ :

$f_Y(y)=\dfrac{\mathrm d}{\mathrm dy}F_X(\ln y)=\dfrac1yf_X(\ln y)=\begin{cases}\frac1{y\sqrt{2\pi}}e^{-\frac12(\ln y)^2}&\text{for }y>0\\0&\text{otherwise}\end{cases}$

The $n$ th moment of $Y$ is

$E[Y^n]=\displaystyle\int_{-\infty}^\infty y^nf_Y(y)\,\mathrm dy=\frac1{\sqrt{2\pi}}\int_0^\infty y^{n-1}e^{-\frac12(\ln y)^2}\,\mathrm dy$

Let $u=\ln y$ , so that $\mathrm du=\frac{\mathrm dy}y$ and $y^n=e^{nu}$ :

$E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu}e^{-\frac12u^2}\,\mathrm du=\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{nu-\frac12u^2}\,\mathrm du$

Complete the square in the exponent:

$nu-\dfrac12u^2=-\dfrac12(u^2-2nu+n^2-n^2)=\dfrac12n^2-\dfrac12(u-n)^2$

$E[Y^n]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{\frac12(n^2-(u-n)^2)}\,\mathrm du=\frac{e^{\frac12n^2}}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du$

But $\frac1{\sqrt{2\pi}}e^{-\frac12(u-n)^2}$ is exactly the PDF of a normal distribution with mean $n$ and variance 1; in other words, the 0th moment of a random variable $U\sim N(n,1)$ :

$E[U^0]=\displaystyle\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac12(u-n)^2}\,\mathrm du=1$

so we end up with

$E[Y^n]=e^{\frac12n^2}$

3 0

3 years ago