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3 years ago

Simplify 6x^2/ cubed root of 3x

Mathematics

1 answer:

hammer [34]3 years ago

4 0

Answer:

6x^2/(3x)^1/3

6= 3 .2

So, 3.2 (x^2)/( 3^1/3) x^1/3

=3^2/3)(2). x^5/3

={ 2 .( 3)^2/3 } x^5/3

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6/42 in simplest form

Ganezh [65]

6/42 in simplest form :

Divide both the nominator and the denominator by the HCF (highest common factor) : 6

6/42 = 6:6/42:6 = 1/7

6 0

2 years ago

Write y=-3x^2-18x-31 in vertex form

Ahat [919]

-3(x+(3))^2-4 is the answer

8 0

3 years ago

Read 2 more answers

I could use a hand here

worty [1.4K]

Answer:

A. 48

B. 24

C. C

Step-by-step explanation:

To divide up whole numbers by fractions, you simply need to just multiply the whole number by the reciprocal of the fraction. in other words, flip the fraction.

Ex.

12 ÷ 1/4

12 x 4

Ex. 2

12 ÷ 1/2

12 x 2

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2 years ago

Read 2 more answers

A personnel manager is concerned about absenteeism. She decides to sample employee records to determine if absenteeism is distri

nlexa [21]

Answer:

$df=categor-1=6-1=5$

The critical value can be founded with the following Excel formula:

=CHISQ.INV(1-0.05,5)

And we got $\chi^2_{critc}= 11.0705$

a. 11.070

And since our calculated value is lower than the critical we FAIL to reject the null hypothesis at 5% of significance

Step-by-step explanation:

Previous concepts

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Solution to the problem

For this case we want to test:

H0: Absenteeism is distributed evenly throughout the week

H1: Absenteeism is NOT distributed evenly throughout the week

We have the following data:

Monday Tuesday Wednesday Thursday Friday Saturday Total

12 9 11 10 9 9 60

The level of significance assumed for this case is $\alpha=0.05$

The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{60}{6}= 10$ and the expected value is the same for all the days since that's what we want to test.

now we can calculate the statistic:

$\chi^2 = \frac{(12-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(11-10)^2}{10}+\frac{(10-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(9-10)^2}{10}=0.8$

Now we can calculate the degrees of freedom (We know that we have 6 categories since we have information for 6 different days) for the statistic given by:

$df=categor-1=6-1=5$

The critical value can be founded with the following Excel formula:

=CHISQ.INV(1-0.05,5)

And we got $\chi^2_{critc}= 11.0705$

a. 11.070

And since our calculated value is lower than the critical we FAIL to reject the null hypothesis at 5% of significance

4 0

3 years ago

Applying Rate of change

olga nikolaevna [1]

$\bf \begin{array}{ccllll}
minutes(x)&Mbs\ left(y)\\
-----&-----\\
2&38.4\\
8&9.6
\end{array}\\\\
-----------------------------\\\\
slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\impliedby \textit{rate of change}
\\\\\\
m=\cfrac{9.6-38.4}{8-2}\implies m=-\cfrac{24}{5}\iff m= -4.8$

now, notice, the rate of change for downloading is negative, because, "y" is decreasing, namely the Mbs to be downloaded, are less and less and less as the minutes go by, because the file is almost fully downloaded

so is -4.8

now... at 8minutes, there are 9.6Mbs to download
bear in mind that 9.6 is just 4.8 * 2
that simply means, another minute, another 4.8 Mbs, and another minute and another 4.8 Mbs and the file is done

so, the file downloaded really in 10minutes

now, we know the rate is -4.8 or -24/5, let us nevermind the sign for now

since we know the file is downloading at 24Mbs in 5minutes, a rate of 24/5

how much is it for 10 minutes? well 10 is really just 5 * 2

so if it downloads at a rate of 24Mbs per 5mins, in 10 minutes it downloaded 24*2 or 48Mbs

7 0

3 years ago