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Bess [88]
3 years ago
14

Find the answer in simplest form 3/7x2/3

Mathematics
1 answer:
Grace [21]3 years ago
8 0

Answer:

7 6/7 or 7.8571429

Step-by-step explanation:

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Answer:

B. An athlete does 10 push-ups <u>for every</u> mile she runs.

Step-by-step explanation:

10/1  ->  20/2  ->  30/3

4 0
3 years ago
Help me please I need good grades!
Andrew [12]
Its yes. 17 over 3 equals to 68 over 12.
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The ratio of girls to boys on the Husky
NemiM [27]

Answer:

63

Step-by-step explanation:

2+3=5

105÷5=21

21×3=63

4 0
3 years ago
The weights of ice cream cartons are normally distributed with a mean weight of 1313 ounces and a standard deviation of 0.50.5 o
asambeis [7]

Answer:

a) We are interested on this probability

P(X>13.17)

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(X>13.17)=P(\frac{X-\mu}{\sigma}>\frac{13.17-\mu}{\sigma})=P(Z>\frac{13.17-13}{0.5})=P(z>0.34)=0.367

b) P(\bar X >13.17)

The new z score is given by:

z =\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And replacing we got:

P(\bar X >13.17)=P(Z> \frac{13.17-13}{\frac{0.5}{\sqrt{36}}})= P(Z>2.04) =0.0207

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Part a

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

X \sim N(13,0.5)  

Where \mu=13 and \sigma=0.5

We are interested on this probability

P(X>13.17)

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(X>13.17)=P(\frac{X-\mu}{\sigma}>\frac{13.17-\mu}{\sigma})=P(Z>\frac{13.17-13}{0.5})=P(z>0.34)=0.367

Part b

Since the distribution for X is normal then we can conclude that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

And we want this probability:

P(\bar X >13.17)

The new z score is given by:

z =\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And replacing we got:

P(\bar X >13.17)=P(Z> \frac{13.17-13}{\frac{0.5}{\sqrt{36}}})= P(Z>2.04) =0.0207

3 0
4 years ago
Help pleaseeeeeee!!!!!
Alexeev081 [22]

Answer:

17

Step-by-step explanation:

4 0
3 years ago
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