Please help! Will give brainliest to correct answer! (1/3) - 50 POINTS - please no wrong answers.

Mathematics

2 answers:

astraxan [27]3 years ago

7 0

Answer:

( 6, pi/6)

Step-by-step explanation:

( 3 sqrt(3), 3)

To get r we use x^2 + y ^2 = r^2

( 3 sqrt(3) )^2 + 3^2 = r^2

9 *3 +9 = r^2

27+9 = r^2

36 = r^2

Taking the square root of each side

sqrt(36) = sqrt(r^2)

6 =r

Now we need to find theta

tan theta = y/x

tan theta = 3 / 3 sqrt(3)

tan theta = 1/ sqrt(3)

Taking the inverse tan of each side

tan ^-1 ( tan theta) = tan ^ -1 ( 1/ sqrt(3))

theta = pi /6

Masteriza [31]3 years ago

7 0

R=sqrt(x^2 + y^2)= sqrt(36)= 6
tan(theta)=y/x= sqrt(3)/3
theta = arctan(sqrt(3)/3)= pi/6

So D) (6, pi/6)

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The service life of a battery used in a cardiac pacemaker is assumed to be normally distributed. A random sample of 10 batteries

Nat2105 [25]

Answer:

(1) We are conducting the one-sample t-test and will be testing for the hypothesis of mean battery greater than 25 h. So, we will reject the null hypothesis for mean battery life equal to 25 hr against the alternative hypothesis for mean battery life greater than 25 hr.

(2) A 95% lower confidence interval on mean battery life is 25.06 hr.

Step-by-step explanation:

We are given that a random sample of 10 batteries is subjected to an accelerated life test by running them continuously at an elevated temperature until failure, and the following lifetimes (in hours) are obtained: 25.5, 26.1, 26.8, 23.2, 24.2, 28.4, 25.0, 27.8, 27.3, and 25.7.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean BAC = $\frac{\sum X}{n}$ = 26 hr

s = sample standard deviation = $\sqrt{\frac{\sum(X - \bar X)^{2} }{n-1} }$ = 1.625 hr

n = sample of batteries = 10

$\mu$ = population mean battery life

Here for constructing a 95% lower confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

(1) It is stated that the manufacturer wants to be certain that the mean battery life exceeds 25 h.

Since we are conducting the one-sample t-test and will be testing for the hypothesis of mean battery greater than 25 h. So, we will reject the null hypothesis for mean battery life equal to 25 hr against the alternative hypothesis for mean battery life greater than 25 hr.

(2) So, a 95% lower confidence interval for the population mean, $\mu$ is;

P(-1.833 < $t_9$ ) = 0.95 {As the lower critical value of t at 9