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3 years ago

Write 3.088 as a mixed number and improper fraction

Mathematics

1 answer:

mixas84 [53]3 years ago

8 0

Answer:

Mixed number-3 and 88/1000 which is simplified to 3 and 11/125

Improper fraction- 386/125

Step-by-step explanation:

<u>Mixed Number</u>

3.088 is read as Three and eighty-eight thousandths. As a fraction we write this as 3 and 88/1000. However, 88/1000 can be simplified. The greatest common factor that will go into 88 and 1000 is 8, so I divide 8 into 88 and get 11, then I divide 8 into 1000 and get 125. Simplified my answer is 3 and 11/125.

<u>Improper Fraction</u>

If I have 3 and 11/125, I will change it to an improper fraction by first multiplying the whole number 3 by the denominator 125. This gives me 375, which I add to the numerator 11 and that gives me 386. I will keep my denominator so my improper fraction will be 386/125.

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How to solve this? Is the question correct? ty :)

bezimeni [28]

Answer:

The proposition is true.

Step-by-step explanation:

Now we proceed to demostrate that expression given is true by algebraic means:

1) $\frac{x^{-1}+y^{-1}}{x^{-1}}+\frac{x^{-1}-y^{-1} }{y^{-1}}$ Given

2) $\frac{y^{-1}\cdot (x^{-1}+y^{-1})+x^{-1}\cdot (x^{-1}-y^{-1})}{x^{-1}\cdot y^{-1}}$ $\frac{a}{b} + \frac{c}{d} = \frac{a\cdot d+b\cdot c}{b\cdot d}$

3) $\frac{x^{-1}\cdot y^{-1}+y^{-2}+x^{-2}-x^{-1}\cdot y^{-1}}{x^{-1}\cdot y^{-1}}$ Distributive property/ $a^{b}\cdot a^{c} = a^{b+c}$

4) $\frac{x^{-2}+y^{-2}}{(x\cdot y)^{-1}}$ Commutative, associative and modulative properties/Existence of additive inverse/ $a^{b}\cdot c^{b} = (a\cdot c)^{b}$

5) $[(x\cdot y)^{-1}]^{-1}\cdot (x^{-2}+y^{-2})$ Commutative property/Definition of division

6) $(x\cdot y)\cdot (x^{-2}+y^{-2})$ $(x^{-1})^{-1}$

7) $x^{-1}\cdot y + x\cdot y^{-1}$ Distributive property/Associative property/ $a^{b}\cdot a^{c} = a^{b+c}$

8) $(x^{-1}\cdot y^{-1})\cdot y^{2} + (x^{-1}\cdot y^{-1})\cdot x^{2}$ Modulative property/Existence of additive inverse/ $a^{b}\cdot a^{c} = a^{b+c}$

9) $(x^{-1}\cdot y^{-1})\cdot (x^{2}+y^{2})$ Distributive property

10) $(x\cdot y)^{-1}\cdot (x^{2}+y^{2})$ $a^{b}\cdot c^{b} = (a\cdot c)^{b}$

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4 0

3 years ago

Given a normal population whose mean is 675 and whose standard deviation is 44, find each of the following: A. The probability t

NNADVOKAT [17]

Answer:

27.88% probability that a random sample of 5 has a mean between 677 and 693.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 675, \sigma = 44, n = 5, s = \frac{44}{\sqrt{5}} = 19.6774$

The probability that a random sample of 5 has a mean between 677 and 693.

This is the pvalue of Z when X = 693 subtracted by the pvalue of Z when X = 677. So

X = 693

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{693 - 675}{19.6774}$

$Z = 0.91$

$Z = 0.91$ has a pvalue of 0.8186

X = 677

$Z = \frac{X - \mu}{s}$

$Z = \frac{677 - 675}{19.6774}$

$Z = 0.1$

$Z = 0.1$ has a pvalue of 0.5398

0.8186 - 0.5398 = 0.2788

27.88% probability that a random sample of 5 has a mean between 677 and 693.

5 0

4 years ago

What is a31 of the<br> arithmetic sequence for<br> which a5 = 12.4 and<br> ag = : 22.4?

MrMuchimi

The value of a₃₁ of the arithmetic sequence exists 77.4.

<h3>How to find the value of a₃₁ of the arithmetic sequence?</h3>

Given: a₅ = 12.4 and a₉ = : 22.4

For the arithmetic sequence a₁, a₂, a₃, ..., the n-th term exists

where d = common difference

a₅ = 12.4,

a₁ + 4d = 12.4 .........(1)

Because a₉ = 22.4,

a₁ + 8d = 22.4 .........(2)

Subtract (1) from (2), we get

a₁ + 8d - (a₁ + 4d) = 22.4 - 12.4

4d = 10

Dividing throughout by 4, we get

d = 2.5

From (1), we get

a₁ = 12.4 - 4 $*$ 2.5 = 2.4

a₃₁ = 2.4 + 30 $*$ 2.5 = 77.4

Therefore, the correct answer is a₃₁ = 77.4

To learn more about the arithmetic sequence refer to; brainly.com/question/6561461

#SPJ9

8 0

1 year ago

The following items are available for use in a pan balance to measure the weight of a large rock. Which items will yield the gre

weqwewe [10]

The correct option is B.
Pan balance is one of the instruments that are used to measure weights of objects. The instrument is designed in such a way that, it has two pans and a standard weight can be added to one of the pan to determine the weight of the object in the other pan. In the question given above, the pan balance is to be used to measure the weight of a large rock. Salt is the right object to be used on the second pan because it can be packaged in standard weights and it is very light in weight. So, it will be very sensitive to the weight of the rock that is been added on the other pan.

8 0

4 years ago

A computer monitor is a rectangle. The display resolution of Monica’s monitor is 240 pixels by 160 pixels. The display resolutio

Bess [88]

Answer:

Monica's monitor is not similar to Eugene's monitor.

Step-by-step explanation:

It is given that the display resolution of Monica's monitor is 240 pixels by 160 pixels and the display resolution of Eugene's monitor is 320 pixels by 200 pixels.

Ratio of the lengths of Monica's monitor and Eugene's monitor = 240:320

= 3:4

Ratio of the heights of Monica's monitor and Eugene's monitor = 160:200

= 4:5

3:4 ≠ 4:5

Hence, Monica's monitor is not similar to Eugene's monitor.

7 0

3 years ago