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3 years ago

What is the difference between 4 5/7 and 1 3/4?

Mathematics

1 answer:

Bas_tet [7]3 years ago

3 0

Answer:

The difference between 4 5/7 and 1 3/4 is 83/28 or 2 27/28

Step-by-step explanation:

First, to make it easier to subtract, let's change these numbers to improper fractions.

4 5/7 = 33/7

1 3/4 = 7/4

Now, we have to find the LCM (least common multiple) between 4 and 7.

4- 4, 8, 12, 16, 20, 24, 28, 32

7- 7, 14, 21, 28, 35, 42, 49

As you can see, the LCM is 28. So, lets change these fractions to have a denominator of 28.

33*4/ 7*4 = 132/ 28

7*7/ 4*7 = 49/ 28

>>132 - 49 = 83

So, the difference between 4 5/7 and 1 3/4 is 83/28 or 2 27/28

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Answer:

10:21 in the morning

Step-by-step explanation:

8:00-8 hrs=12:00

12:00-1 hr and 39 min = 10:21

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2 years ago

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A random sample of 36 students at a community college showed an average age of 25 years. Assume the ages of all students at the

Pavel [41]

Answer:

98% confidence interval for the average age of all students is [24.302 , 25.698]

Step-by-step explanation:

We are given that a random sample of 36 students at a community college showed an average age of 25 years.

Also, assuming that the ages of all students at the college are normally distributed with a standard deviation of 1.8 years.

So, the pivotal quantity for 98% confidence interval for the average age is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample average age = 25 years

$\sigma$ = population standard deviation = 1.8 years

n = sample of students = 36

$\mu$ = population average age

So, 98% confidence interval for the average age, $\mu$ is ;

P(-2.3263 < N(0,1) < 2.3263) = 0.98

P(-2.3263 < $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < 2.3263) = 0.98

P( $-2.3263 \times {\frac{\sigma}{\sqrt{n} }$ < ${\bar X - \mu}$ < $2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ) = 0.98

P( $\bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} }$ < $\mu$ < $\bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ) = 0.98

98% confidence interval for $\mu$ = [ $\bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} }$ , $\bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ]

= [ $25 - 2.3263 \times {\frac{1.8}{\sqrt{36} }$ , $25 + 2.3263 \times {\frac{1.8}{\sqrt{36} }$ ]

= [24.302 , 25.698]

Therefore, 98% confidence interval for the average age of all students at this college is [24.302 , 25.698].

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3 years ago

The length of a rectangle is 8cm and the height is 16cm find the answer of the above figures using this formular...

notsponge [240]

Length=8cm
Breadth=16cm(Height)

<h3>Perimeter:-</h3>

$\\ \sf\longmapsto Perimeter=2(L+B)$

$\\ \sf\longmapsto Perimeter=2(16+8)$

$\\ \sf\longmapsto Perimeter=2(24)$

$\\ \sf\longmapsto Perimeter=48cm$

Area:-

$\\ \sf\longmapsto Area=Length\times Breadth$

$\\ \sf\longmapsto Area=8(16)$

$\\ \sf\longmapsto Area=128cm^2$

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3 years ago

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AB = 6.1

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Answer:

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Step-by-step explanation:

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3 years ago