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3 years ago

What is a equivalent ratio to 6/25?

Mathematics

1 answer:

Colt1911 [192]3 years ago

5 0

6 / 25

= 6 / 25 * (2 / 2) = 12 / 50

= 6 / 25 * (3 / 3) = 18 / 75

Equivalent ratio is 12 / 50 or 18 / 75

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you are riding a bicycle which has tires with a 25-inch diameter at a steady 15-miles per hour, what is the angular velocity of

motikmotik

Answer:

The angular velocity is 6.72 π radians per second

Step-by-step explanation:

The formula of the angular velocity is ω = $\frac{v}{r}$ , where v is the linear velocity and r is the radius of the circle

The unit of the angular velocity is radians per second

∵ The diameter of the tire is 25 inches

∵ The linear velocity is 15 miles per hour

- We must change the mile to inch and the hour to seconds

∵ 1 mile = 63360 inches

∵ 1 hour = 3600 second

∴ 15 miles/hour = 15 × $\frac{63360}{3600}$

∴ 15 miles/hour = 264 inches per second

Now let us find the angular velocity

∵ ω = $\frac{v}{r}$

∵ v = 264 in./sec.

∵ d = 25 in.

- The radius is one-half the diameter

∴ r = $\frac{1}{2}$ × 25 = 12.5 in.

- Substitute the values of v and r in the formula above to find ω

∴ ω = $\frac{264}{12.5}$

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2 years ago

Assume a simple random sample of 10 BMIs with a standard deviation of 1.186 is selected from a normally distributed population o

kirza4 [7]

Answer:

a) H0: $\sigma = 1.34$

H1: $\sigma \neq 1.34$

b) $df = n-1= 10-1=9$

And the critical values with $\alpha/2=0.005$ on each tail are:

$\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589$

c) $t=(10-1) [\frac{1.186}{1.34}]^2 =7.05$

d) For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

Step-by-step explanation:

Information provided

n = 10 sample size

s= 1.186 the sample deviation

$\sigma_o =1.34$ the value that we want to test

$p_v$ represent the p value for the test

t represent the statistic (chi square test)

$\alpha=0.01$ significance level

Part a

On this case we want to test if the true deviation is 1,34 or no, so the system of hypothesis are:

H0: $\sigma = 1.34$

H1: $\sigma \neq 1.34$

The statistic is given by:

$t=(n-1) [\frac{s}{\sigma_o}]^2$

Part b

The degrees of freedom are given by:

$df = n-1= 10-1=9$

And the critical values with $\alpha/2=0.005$ on each tail are:

$\chi_{\alpha/2}= 1.735, \chi_{1-\alpha/2}= 23.589$

Part c

Replacing the info we got:

$t=(10-1) [\frac{1.186}{1.34}]^2 =7.05$

Part d

For this case since the critical value is not higher or lower than the critical values we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not significantly different from 1.34

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2 years ago