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3 years ago

Find radius of a circle whose arclength is 57.1 meters and central angle of 1.46 radians

1 answer:

max2010maxim [7]3 years ago

3 0

Hint: $\bf s=r\theta\implies \cfrac{s}{\theta}=r\qquad 
\begin{cases}
s=\textit{arc's length}\\
\theta=\textit{central angle, in radians}\\
r=radius
\end{cases}$

You might be interested in

A university wants to compare out-of-state applicants' mean SAT math scores (?1) to in-state applicants' mean SAT math scores (?

nordsb [41]

Answer:

d. Yes, because the confidence interval does not contain zero.

Step-by-step explanation:

We are given that the university looks at 35 in-state applicants and 35 out-of-state applicants. The mean SAT math score for in-state applicants was 540, with a standard deviation of 20.

The mean SAT math score for out-of-state applicants was 555, with a standard deviation of 25.

Firstly, the Pivotal quantity for 95% confidence interval for the difference between the population means is given by;

P.Q. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ~ $t__n__1-_n__2-2$

where, $\bar X_1$ = sample mean SAT math score for in-state applicants = 540

$\bar X_2$ = sample mean SAT math score for out-of-state applicants = 555

$s_1$ = sample standard deviation for in-state applicants = 20

$s_2$ = sample standard deviation for out-of-state applicants = 25

$n_1$ = sample of in-state applicants = 35

$n_2$ = sample of out-of-state applicants = 35

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2} +(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(35-1)\times 20^{2} +(35-1)\times 25^{2} }{35+35-2} }$ = 22.64

Here for constructing 95% confidence interval we have used Two-sample t test statistics.

So, 95% confidence interval for the difference between population means ( $\mu_1-\mu_2$ ) is ;

P(-1.997 < $t_6_8$ < 1.997) = 0.95 {As the critical value of t at 68 degree

of freedom are -1.997 & 1.997 with P = 2.5%}

P(-1.997 < $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < 1.997) = 0.95

P( $-1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < ${(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}$ < $1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ) = 0.95

P( $(\bar X_1-\bar X_2)-1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < ( $\mu_1-\mu_2$ ) < $(\bar X_1-\bar X_2)+1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ) = 0.95

95% confidence interval for ( $\mu_1-\mu_2$ ) =

[ $(\bar X_1-\bar X_2)-1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ , $(\bar X_1-\bar X_2)+1.997 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ]

=[ $(540-555)-1.997 \times {22.64 \times \sqrt{\frac{1}{35} +\frac{1}{35} } }$ , $(540-555)+1.997 \times {22.64 \times \sqrt{\frac{1}{35} +\frac{1}{35} } }$ ]

= [-25.81 , -4.19]

Therefore, 95% confidence interval for the difference between population means SAT math score for in-state and out-of-state applicants is [-25.81 , -4.19].

This means that the mean SAT math scores for in-state students and out-of-state students differ because the confidence interval does not contain zero.

So, option d is correct as Yes, because the confidence interval does not contain zero.

6 0

3 years ago

Donald is 38 years younger than Natalie. 5 years ago, Natalie's age was 3 times Donald's age. How old is Donald now?

Svetlanka [38]

9514 1404 393

Answer:

Step-by-step explanation:

Let d represent Donald's age now. Then Natali's age now is d+38. The age relationship 5 years ago was ...

(d+38) -5 = 3(d -5)

d +33 = 3d -15 . . . . . eliminate parentheses

48 = 2d . . . . . . . . . . . add 15-d

24 = d . . . . . . . . . . . . . divide by 2

Donald is 24 years old now.

_____

Alternate solution

You can also work this by considering that 5 years ago, Natalie's age was more than Donald's age by twice Donald's age then. Hence Donald was 38/2 = 19 at that time. Now, Donald is 19+5 = 24.

4 0

3 years ago

Write --X2 = 15 in standard form.

drek231 [11]

I’m assuming this is -x^2 = 15
standard form: -x^2 - 15 = 0

8 0

3 years ago

Which complex number is a distance of sqrt17 from the origin on the complex plane?

lions [1.4K]

Let z=x+iy be a complex number. The distance from this complex number to the origin is

$\sqrt{(x-0)^2+(y-0)^2}=\sqrt{x^2+y^2}.$