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stiv31 [10]
4 years ago
5

Which segment is not a radius of circle C cq cp rc pq

Mathematics
2 answers:
attashe74 [19]4 years ago
7 0

The answer could be D- pq. It's not from the center of the circle.

zubka84 [21]4 years ago
4 0

Answer:

D. PQ.

Step-by-step explanation:

We have been given a circle. We are asked to find the segment which is not a radius of circle C.

We know that radius of a circle is a segment that connects any point on circle to center of the circle.

Upon looking at our given circle, we can see that CQ, CP and RC are radii of the circle C.

Since PQ is the diameter of our given circle, therefore, segment PQ is not radius of our given circle.

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A study was conducted to determine whether magnets were effective in treating pain. The values represent measurements of pain us
gregori [183]

Answer and Step-by-step explanation: The null and alternative hypothesis for this test are:

H_{0}: s_{1}^{2} = s_{2}^{2}

H_{a}: s_{1}^{2} > s_{2}^{2}

To test it, use F-test statistics and compare variances of each treatment.

Calculate F-value:

F=\frac{s^{2}_{1}}{s^{2}_{2}}

F=\frac{1.26^{2}}{0.93^{2}}

F=\frac{1.5876}{0.8649}

F = 1.8356

The <u>critical value of F</u> is given by a F-distribution table with:

degree of freedom (row): 20 - 1 = 19

degree of freedom (column): 20 - 1 = 19

And a significance level: α = 0.05

F_{critical} = 2.2341

Comparing both values of F:

1.856 < 2.2341

i.e. F-value calculated is less than F-value of the table.

Therefore, failed to reject H_{0}, meaning there is <u>no sufficient data to support the claim</u> that sham treatment have pain reductions which vary more than for those using magnets treatment.

4 0
3 years ago
Need answer thanks! only answer if you know please​!
Vaselesa [24]

Answer:

a) k \approx 0.0602\,\frac{1}{min} (k\approx  6.02\,\frac{\%}{min}), b) k \approx 0.0648\,\frac{1}{min} (k \approx 6.48\,\frac{\%}{min}), c) t_{1/2} \approx 60.274\,days, d) t_{1/2}\approx 4.305\,min, e) k \approx 0.3\,\frac{1}{min} (k\approx 30\,\frac{\%}{min}), f) k \approx 0.0120\,\frac{1}{min} (k \approx 1.20\,\frac{\%}{min}), g) k \approx 2.876\times 10^{-5}\,\frac{1}{yr} (k \approx 2.876\times 10^{-3}\,\frac{\%}{yr})

Step-by-step explanation:

All radioactive isotopes decay exponentially, the mass of the isotope as a function of time (m(t)), measured in grams, is defined by:

m(t) = m_{o}\cdot e^{-k\cdot t} (1)

Where:

m_{o} - Initial mass of the isotope, measured in grams.

k - Decay rate, measured in \frac{1}{min} or \frac{1}{day} or \frac{1}{yr}.

t - Time, measured in minutes, days or years.

We define the decay rate in terms of half-life by using the following expression:

k = \frac{\ln 2}{t_{1/2}} (2)

Where t_{1/2} is the half-life of the isotope, measured in minutes or years.

Now we proceed to determine the missing values:

a) Polonium-200 (t_{1/2} = 11.5\,min)

k = \frac{\ln 2}{11.5\,min}

k \approx 0.0602\,\frac{1}{min}

b) Lead-194 (t_{1/2} = 10.7\,min)

k = \frac{\ln 2}{10.7\,min}

k \approx 0.0648\,\frac{1}{min}

c) Iodine-125 (k = 0.0115\,\frac{1}{day})

t_{1/2} = \frac{\ln 2}{k}

t_{1/2} = \frac{\ln 2}{0.0115\,\frac{1}{day} }

t_{1/2} \approx 60.274\,days

d) Kryption-75 (k = 0.161\,\frac{1}{min})

t_{1/2} = \frac{\ln 2}{k}

t_{1/2} = \frac{\ln 2}{0.161\,\frac{1}{min} }

t_{1/2}\approx 4.305\,min

e) Strontium-79 (t_{1/2} = 2.3\,min)

k = \frac{\ln 2}{2.3\,min}

k \approx 0.3\,\frac{1}{min}

f) Uranium-229 (t_{1/2} = 58\,min)

k = \frac{\ln 2}{58\,min}

k \approx 0.0120\,\frac{1}{min}

g) Plutonium-239 (t_{1/2} = 24100\,yr)

k = \frac{\ln 2}{24100\,yr}

k \approx 2.876\times 10^{-5}\,\frac{1}{yr}

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3 years ago
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Can’t do 16 but here are the rest :))

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8,-4 : 6,-2 : 5,-1 All should add up to four.
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PLZZZZZZZZZ NEED HELP ASAP QUESTION WORTH 20 POINTS
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Part A: After 9 days the radius of algae was approximately 12.81 mm. The reasonable domain to point is (0,9).

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3 years ago
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