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3 years ago

How do I solve 6/2 + 8/3

Mathematics

1 answer:

Xelga [282]3 years ago

7 0

Answer:

The solved given fractional expression is $\frac{17}{3}$

Therefore the given fractional expression becomes

$\frac{6}{2}+\frac{8}{3}=\frac{17}{3}$

Step-by-step explanation:

Given fractional expression is $\frac{6}{2}+\frac{8}{3}$

To solve the given fractional expression as below :

$\frac{6}{2}+\frac{8}{3}$

$=3+\frac{8}{3}$

$=\frac{3\times 3+8}{3}$

( here taking the term 3 as LCM )

$=\frac{9+8}{3}$ ( by adding the sums here )

$=\frac{17}{3}$

$\frac{6}{2}+\frac{8}{3}=\frac{17}{3}$

Therefore the solved given fractional expression is $\frac{17}{3}$

Therefore the given fractional expression becomes

$\frac{6}{2}+\frac{8}{3}=\frac{17}{3}$

The solved given fractional expression is $\frac{6}{2}+\frac{8}{3}=\frac{17}{3}$

You might be interested in

Which rule represents the table's values?<br> y = 2x<br> y=-2x<br> y=-12%<br> V = -2x + 1

hram777 [196]

Answer:

Therefore the required rule for the given table is

$y=-2x$

Step-by-step explanation:

Given :

Table values are

x y

6 -12

7 -14

8 -16

9 -18

10 -20

To Find :

The rule for above result = ?

Solution:

The Required Rule for the above Table is

$y=-2x$

For first case

Put x = 6 in the rule we get

$y=-2\times 6 =-12$ as shown in the table

For second case

Put x = 7 in the rule we get

$y=-2\times 7 =-14$ as shown in the table

For third case

Put x = 8 in the rule we get

$y=-2\times 8 =-16$ as shown in the table

For fourth case

Put x = 9 in the rule we get

$y=-2\times 9 =-18$ as shown in the table

For fifth case

Put x = 10 in the rule we get

$y=-2\times 10 =-20$ as shown in the table

Therefore the required rule for the given table is

$y=-2x$

7 0

4 years ago

Sick-leave time used by employees of a firm in a course of one month has approximately normal distribution, with a mean of 200 h

Usimov [2.4K]

Answer:

a)0.62% probability that total sick leave for next month will be less than 150 hours.

b) 225.6 hours should be budgeted for sick leave if that amount is to be exceeded with a probability of only 0.10.

Step-by-step explanation:

Problems of normally distributed samples can be 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.

In this problem, we have that:

$\mu = 200, \sigma = \sqrt{400} = 20$

a.Find the probability that total sick leave for next month will be less than 150 hours.

This probability is the pvalue of Z when X = 150. So:

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

$Z = \frac{150 - 200}{20}$

$Z = -2.5$

$Z = -2.5$ has a pvalue of 0.0062.

So there is a 0.62% probability that total sick leave for next month will be less than 150 hours.

b.In planning schedules for next month, how much time should be budgeted for sick leave if that amount is to be exceeded with a probability of only 0.10.

This is the value of X when Z has a pvalue of 0.90. So $Z = 1.28$