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Luba_88 [7]
3 years ago
13

Which of the following is NOT an example of matter?

Mathematics
1 answer:
klio [65]3 years ago
6 0
The answer is sunlight :)
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A fair number cube is labeled with the numbers 1 - 6. After rolling the cube 100 times, which
Kipish [7]

Answer:

NONE. You loose. The question says a “FAIR” number cube, so by definition, NONE of the sides comes up MOST often. It’s an invalid QUESTION! Indeterminate Answer.

If it’s a FAIR die, then any of the 6 numbers has the same chance of coming up: 1/6

So, there is no MOST LIKELY result for 100 rolls.

Step-by-step explanation:

3 0
2 years ago
Help needed asap please?
OleMash [197]

Answer:

its 72

Step-by-step explanation:

i already solved this question yesterday. :)

6 0
3 years ago
Indicate the equation of the given line in standard form. The line with slope and containing the midpoint of the segment whose e
Tems11 [23]

Answer:

x + y = -1

Step-by-step explanation:

To write the equation of a line, find the slope and use it with a point in the point slope form.

You can find the slope using the slope formula.

m = \frac{y_2-y_1}{x_2-x_1} = \frac{5--3}{-6-2} = \frac{8}{-8}=-1

Substitute m = -1 and the point (2,-3) into the point slope form.

y - y_1 = m(x-x_1)\\y --3 = -1(x-2)\\y+3=-1(x-2)

Convert to standard form by applying the distributive property and rearranging the terms.

y+3 = -1(x-2)

y + 3 = -x + 2

x + y + 3 = 2

x + y = -1

8 0
3 years ago
If f(x)=3x5−10x3+5, on what intervals is f″(x)>0?
Ksivusya [100]

Answer:

(0.7937,\infty)

Step-by-step explanation:

we are given an algebraic function in x, such that

f(x) = 3x^5-10x^3+5

To find when second derivative >0

For this let us find first and second derivative by successive differentiation.

f(x) = 3x^5-10x^3+5\\f'(x) =15x^4-30x^2\\f'(x)=60x^3-30

when ii derivative >0 we have

60x^3-30>0

x^3>0.5\\x>0.7937

f"(x)>0 in

(0.7937,\infty)

5 0
2 years ago
An article includes the accompanying data on compression strength (lb) for a sample of 12-oz aluminum cans filled with strawberr
Veseljchak [2.6K]

Answer:

<em>The calculated value |t| = 2.375 > 2.1009  at 0.05 level of significance</em>

<em>Null hypothesis is rejected </em>

<em>Alternative hypothesis is accepted</em>

<em>The extra carbonation of cola results in a higher average compression strength</em>

<u><em /></u>

Step-by-step explanation:

<u><em>Step(i):-</em></u>

<em>Given data</em>

<em>First sample size (n₁) = 10 </em>

<em>mean of the first sample(x₁⁻) = 537</em>

<em>standard deviation of the first sample (S₁) = 22</em>

<em>second sample size (n₂) = 10 </em>

<em>mean of the second sample (x₂⁻) = 559</em>

<em>standard deviation of the second sample (S₂) = 17</em>

<u><em>Step(ii)</em></u><em>:-</em>

<u><em>Null hypothesis : H₀:</em></u><em>- The extra carbonation of cola results in a lower average compression strength</em>

<u>Alternative Hypothesis :H₁</u>

<em>The extra carbonation of cola results in a higher average compression strength</em>

<u><em>Step(iii)</em></u><em>:-</em>

<em>By using student's t -test for difference of means</em>

<u><em>Test statistic</em></u>

<em>       </em>t = \frac{x^{-} _{1}-x^{-} _{2}  }{\sqrt{S^{2} (\frac{1}{n_{1} }+\frac{1}{n_{2} }  } )}<em />

<em>  where </em>

<em>     </em>S^{2}  = \frac{n_{1}S^{2} _{1} + n_{2} S_{2} ^{2}  }{n_{1}+n_{2} -2 }<em />

<em>    </em>S^{2}  = \frac{10(22)^{2}  + 10 (17) ^{2}  }{10+10 -2 } = \frac{ 7730}{18} = 429.4<em />

<em>    </em>t = \frac{537-559 }{\sqrt{429.4 (\frac{1}{10 }+\frac{1}{10 }  } )}<em />

<em>   t =  -2.375</em>

<em>|t| = |-2.375| = 2.375</em>

<em>Degrees of freedom</em>

<em>γ = n₁+n₂ -2 = 10+10-2 =18</em>

<em />t_{\frac{\alpha }{2} ,n-1}=t_{(\frac{0.05}{2} ,18)} = t_{(0.025,18)}}=2.1009<em />

<em>The calculated value |t| = 2.375 > 2.1009  at 0.05 level of significance</em>

<em>Null hypothesis is rejected </em>

<em>Alternative hypothesis is accepted</em>

<u><em>Final answer:-</em></u>

<em>The extra carbonation of cola results in a higher average compression strength</em>

<u><em /></u>

<em />

<em> </em>

<em />

5 0
3 years ago
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