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r-ruslan [8.4K]
3 years ago
15

Work out 18/36 - 2/6, give your answer in its simplest form​

Mathematics
1 answer:
svp [43]3 years ago
4 0
The answer in simplest form would be 1/6
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Compare the modeld for one third and two sixths
DochEvi [55]
1/3 is equal to 2/6
This is because 1/3 converted to 6ths you have to multiply 1/3 times 2 which equals 2/6
1/3 = 2/6
4 0
3 years ago
Express 36 mins as a fraction of an hour​
Oxana [17]

Answer:

36/60

Step-by-step explanation:

There are 60 minutes in an hour

7 0
2 years ago
Katie gives Maggie half of her pencils. Maggie keeps 5 and gives the rest to jamil. Write an expression for the number of pencil
Inga [223]

Answer

Write an expression for the number of pencils Maggie gives to jamil.

To prove

Let us assume that the number of pencils Katie have = x

As given

Katie gives Maggie half of her pencils.

Katie\ gives\ Maggie\ half\ of\ her\ pencils = \frac{1x}{2}

As given

Maggie keeps 5 and gives the rest to jamil.

Thus

Number\ of\ pencils\ Maggie\ gives\ to\ jamil = \frac{1x}{2} - 5

Therefore the expression for the number of pencils Maggie gives to jamil are  \frac{1x}{2} - 5.





5 0
2 years ago
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m_a_m_a [10]
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7 0
3 years ago
Thompson and Thompson is a steel bolts manufacturing company. Their current steel bolts have a mean diameter of 144 millimeters,
valentinak56 [21]

Answer:

The probability that the sample mean would differ from the population mean by more than 2.6 mm is 0.0043.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample  means will be approximately normally distributed.

Then, the mean of the distribution of sample mean is given by,

\mu_{\bar x}=\mu

And the standard deviation of the distribution of sample mean  is given by,

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}

The information provided is:

<em>μ</em> = 144 mm

<em>σ</em> = 7 mm

<em>n</em> = 50.

Since <em>n</em> = 50 > 30, the Central limit theorem can be applied to approximate the sampling distribution of sample mean.

\bar X\sim N(\mu_{\bar x}=144, \sigma_{\bar x}^{2}=0.98)

Compute the probability that the sample mean would differ from the population mean by more than 2.6 mm as follows:

P(\bar X-\mu_{\bar x}>2.6)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}} >\frac{2.6}{\sqrt{0.98}})

                           =P(Z>2.63)\\=1-P(Z

*Use a <em>z</em>-table for the probability.

Thus, the probability that the sample mean would differ from the population mean by more than 2.6 mm is 0.0043.

8 0
3 years ago
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