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2 years ago

Moira solved rubik's cube in 3 minutes and 30 seconds. what is moira's solving rate?

1 answer:

7 0

In the context of rubiks cube solving the rate at which one can solve a rubiks cube is determined by the number of cubes solved and the time taken

In this question we were told that moira solved a rubiks cube in 3 minutes 30 seconds

Let us convert time total time to seconds

3 minutes 30 in seconds = 3*60+ 30 = 180+30 = 210 seconds

Hence Moira rate = 1 rubiks cube/210 seconds

Learn more about rate:

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A number decreased by 23 is 44. Find the number.

Irina-Kira [14]

A) can.
because the number (x) has been decreased by 23 and is now 44.

Hope this helps :)

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2 years ago

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F(x)= 2x^3- x^2 +x+ 1 is divided by 2x +1.

ololo11 [35]

Answer:

Step-by-step explanation:

x^2

--------------------------------------------------

2x + 1 / 2x^3 - x^2 + x + 1

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2 years ago

Millie works 8 hours on Saturday and earns £6.50 an hour. Her bus fare to work costs £2.75 and she also spends £4.25 on her lunc

Nimfa-mama [501]

The answer is 7.50 ......

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2 years ago

When solving subtraction problems, what should you do when there aren't enough counters to remove?

OleMash [197]

Answer: If you are little you can use you fingers but if you are older you can try to write it down.

Step-by-step explanation:

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2 years ago

Give a 98% confidence interval for one population mean, given asample of 28 data points with sample mean 30.0 and sample standar

klio [65]

We have a sample of 28 data points. The sample mean is 30.0 and the sample standard deviation is 2.40. The confidence level required is 98%. Then, we calculate α by:

$\begin{gathered} 1-\alpha=0.98 \\ \alpha=0.02 \end{gathered}$

The confidence interval for the population mean, given the sample mean μ and the sample standard deviation σ, can be calculated as:

$CI(\mu)=\lbrack x-Z_{1-\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}},x+Z_{1-\frac{\alpha}{2}}\cdot\frac{\sigma}{\sqrt[]{n}}\rbrack$

Where n is the sample size, and Z is the z-score for 1 - α/2. Using the known values:

$CI(\mu)=\lbrack30.0-Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}},30.0+Z_{0.99}\cdot\frac{2.40}{\sqrt[]{28}}\rbrack$

Where (from tables):

$Z_{0.99}=2.33$

Finally, the interval at 98% confidence level is:

$CI(\mu)=\lbrack28.94,31.06\rbrack$

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6 months ago