Which number expresses 6.72 as a fraction in simplest form?

Mathematics

1 answer:

Dimas [21]3 years ago

5 0

The number that expresses 6.72 as a fraction is simplest form is...

6.72= 168/25

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A frequency table of grades has five classes (A, B, C, D,F) with frequencies of 4, 10, 17, 6, and respectively. Using percentage

fiasKO [112]

We have the frequencies for each of the grades. We can estimate the number of students graded by adding all those frequencies. Let's call N the total number of grades:

$\begin{gathered} N=4+10+17+6+2 \\ N=39 \end{gathered}$

We have then a total number of grades of 39.

The corresponding relative frequency for a grade is the ratio of the frequency to the total number of "samples", 39 in this case.

Then, for grade A, the relative frequency (RF) will be:

$RF_A=\frac{4}{39}\approx0.10256\approx10.26\text{ \%}$

This will be the fraction of the total grades that are A. Represented as a percentage will be 10.26%, rounded to two decimal places.

Now, to complete the table we do the same for the other frequencies:

For grade B:

$RF_B=\frac{10}{39}\approx0.2564\approx25.64\text{ \%}$

For grade C:

$RF_C=\frac{17}{39}\approx0.4359\approx43.59\text{ \%}$

For grade D:

$RF_D=\frac{6}{39}\approx0.1538\approx15.38\text{ \%}$

For grade F:

$RF_D=\frac{2}{39}\approx0.0513\approx5.13\text{ \%}$

4 0

1 year ago

a slitter assembly contains 48 blades five blades are selected at random and evaluated each day for sharpness if any dull blade

son4ous [18]

Answer:

P(at least 1 dull blade)=0.7068

Step-by-step explanation:

I hope this helps.

This is what it's called dependent event probability, with the added condition that at least 1 out of 5 blades picked is dull, because from your selection of 5, you only need one defective to decide on replacing all.

So if you look at this from another perspective, you have only one event that makes it so you don't change the blades: that 5 out 5 blades picked are sharp. You also know that the probability of changing the blades plus the probability of not changing them is equal to 100%, because that involves all the events possible.

P(at least 1 dull blade out of 5)+Probability(no dull blades out of 5)=1

P(at least 1 dull blade)=1-P(no dull blades)

But the event of picking one blade is dependent of the previous picking, meaning there is no chance of picking the same blade twice.

So you have 38/48 on getting a sharp one on your first pick, then 37/47 (since you remove 1 sharp from the possibilities, and 1 from the whole lot), and so on.

Also since are consecutive events, you need to multiply the events.

The probability that the assembly is replaced the first day is:

P(at least 1 dull blade)=1-P(no dull blades)

$P(at least 1 dull blade)=1-(\frac{38}{48}* \frac{37}{47} *\frac{36}{46}*\frac{35}{45}*\frac{34}{44})$