What is 96percent as a frction in simplest form?

9.3.2 Listed below are body temperatures from five different subjects measured at 8 AM and again at 12 AM. Find the values of d

kotegsom [21]

Answer:

$\frac{}{d}$ = −0.26

$s_{d}$ = 0.4219

Step-by-step explanation:

Given:

Sample1: 98.1 98.8 97.3 97.5 97.9

Sample2: 98.7 99.4 97.7 97.1 98.0

Sample 1 Sample 2 Difference d

98.1 98.7 -0.6

98.8 99.4 -0.6

97.3 97.7 -0.4

97.5 97.1 0.4

97.9 98.0 -0.1

To find:

Find the values of $\frac{}{d}$ and $s_{d}$

d overbar ( $\frac{}{d}$ ) is the sample mean of the differences which is calculated by dividing the sum of all the values of difference d with the number of values i.e. n = 5

$\frac{}{d}$ = ∑d/n

= (−0.6 −0.6 −0.4 +0.4 −0.1) / 5

= −1.3 / 5

$\frac{}{d}$ = −0.26

s Subscript d is the sample standard deviation of the difference which is calculated as following:

$s_{d}$ = √∑( $d_{i}$ - $\frac{}{d}$ )²/ n-1

$s_{d}$ =

√ $(-0.6 - (-0.26))^{2} + (-0.6 - (-0.26))^{2} + (-0.4 - (-0.26))^{2} + (0.4-(-0.26))^{2} + (-0.1 - (-0.26))^{2} / 5-1$

= √ (−0.6 − (−0.26 ))² + (−0.6 − (−0.26))² + (−0.4 − (−0.26))² + (0.4 −

(−0.26))² + (−0.1 − (−0.26))² / 5−1

= $\sqrt{\frac{0.1156 + 0.1156 + 0.0196 + 0.4356 + 0.0256}{4} }$

= $\sqrt{\frac{0.712}{4} }$

= $\sqrt{0.178}$

= 0.4219

$s_{d}$ = 0.4219

Subscript d represent

μ $_{d}$ represents the mean of differences in body temperatures measured at 8 AM and at 12 AM of population.

3 0

2 years ago

How do I find the arc length with given radius & central angel?

ella [17]

The length of an arc subtended by a central angle is proportional to the circle's circumference. If the given central angle has measure $\theta$ and the circle has radius $r$ , then the arc has length $\ell$ such that

$\dfrac{\ell}\theta=\dfrac{2\pi r}{2\pi}=r$

where $\theta$ is given in radians. Then

$\ell=r\theta$

If $\theta$ is given in degrees, then

$\dfrac{\ell}\theta=\dfrac{2\pi r}{360}\implies\ell=\dfrac{\theta\pi r}{180}$

8 0

3 years ago

Brent is a researcher for a food company. He is on a team creating a reduced-calorie version of its flagship cracker. The team w

Andre45 [30]

Answer:

Step-by-step explanation:

Hello!

The research team created a cracker with fewer calories. The average content of calories of the new crackers per serving of 6 should be less than 60.

To test it a random sample of 26 samples of the new cracker was taken and the calories per serving were measured.

Then the study variable is

X: calories of a 6 serve sample of the new reduced-calorie version. (cal)

The variable has a normal distribution with a population standard deviation of 0.82 cal.

To test the claim that the new crackers have on average less than 60 calories, the parameter of interest is the population mean (μ) and the hypotheses are:

H₀: μ ≥ 60

H₁: μ < 60

α: 0.01

Since the variable has a normal distribution and the population variance is known, the best statistic to use to conduct the test is a Standard Normal

$Z= \frac{(X[bar]-Mu)}{\frac{Sigma}{\sqrt{n}} } ~N(0;1)$