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2 years ago

Simplify fourth root of 6 over fifth root of 6.

Mathematics

2 answers:

grigory [225]2 years ago

8 0

Answer:

$\sqrt[20]{6}$

Step-by-step explanation:

we need to simplify

$\frac{\sqrt[4]{5}}{\sqrt[5]{6}}$

Each radical can be written in fraction form

$\sqrt[4]{6} = 6^{\frac{1}{4}}$

$\sqrt[5]{6} = 6^{\frac{1}{5}}$

$\frac{\sqrt[4]{5}}{\sqrt[5]{6}}=\frac{6^{\frac{1}{4}}}{6^{\frac{1}{5}}}$

Both top and bottom have same base so we subtract the exponents

$6^{\frac{1}{4} -\frac{1}{5}}$

Take common denominator to subtract the fractions

$\frac{1*5}{4*5} -\frac{1*4}{5*4}$

$\frac{5-4}{20} =\frac{1}{20}$

$6^{\frac{1}{4} -\frac{1}{5}}=6^\frac{1}{20}$

$6^\frac{1}{20} =\sqrt[20]{6}$

nikklg [1K]2 years ago

7 0

Simplified form would be twentieth root of 6

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Two soccer teams play 8 games in their season. The number of goals each team scored per game is listed below: Team X: 11, 3, 0,

Gre4nikov [31]

Answer:

C. Team Y’s scores have a lower mean value.

Step-by-step explanation:

We are given that Two soccer teams play 8 games in their season. The number of goals each team scored per game is listed below:

Team X: 11, 3, 0, 0, 2, 0, 6, 4

Team Y: 4, 2, 0, 3, 2, 1, 6, 4

Firstly, we will calculate the mean, median, range and inter-quartile range for Team X;

Mean of Team X data is given by the following formula;

Mean, $\bar X$ = $\frac{\sum X}{n}$

= $\frac{11+ 3+ 0+ 0+ 2+ 0+ 6+ 4}{8}$ = $\frac{26}{8}$ = 3.25

So, the mean of Team X's scores is 3.25.

Now, for calculating the median; we have to arrange the data in ascending order and then observe that the number of observations (n) in the data is even or odd.

Team X: 0, 0, 0, 2, 3, 4, 6, 11

If n is odd, then the formula for calculating median is given by;

Median = $(\frac{n+1}{2} )^{th} \text{ obs.}$

If n is even, then the formula for calculating median is given by;

Median = $\frac{(\frac{n}{2})^{th} \text{ obs.} +(\frac{n}{2}+1)^{th} \text{ obs.} }{2}$

Here, the number of observations is even, i.e. n = 8.

So, Median = $\frac{(\frac{n}{2})^{th} \text{ obs.} +(\frac{n}{2}+1)^{th} \text{ obs.} }{2}$

= $\frac{(\frac{8}{2})^{th} \text{ obs.} +(\frac{8}{2}+1)^{th} \text{ obs.} }{2}$

= $\frac{(4)^{th} \text{ obs.} +(5)^{th} \text{ obs.} }{2}$

= $\frac{2+3}{2}$ = 2.5

So, the median of Team X's score is 2.5.

Now, the range is calculated as the difference between the highest and the lowest value in our data.

Range = Highest value - Lowest value

= 11 - 0 = 11

So, the range of Team X's score is 11.

Now, the inter-quartile range of the data is given by;

Inter-quartile range = $Q_3-Q_1$

$Q_1=(\frac{n+1}{4} )^{th} \text{ obs.}$

= $(\frac{8+1}{4} )^{th} \text{ obs.}$

= $(2.25 )^{th} \text{ obs.}$

$Q_1 = 2^{nd} \text{ obs.} + 0.25[ 3^{rd} \text{ obs.} -2^{nd} \text{ obs.} ]$

= 0 + 0.25[0 - 0] = 0

$Q_3=3(\frac{n+1}{4} )^{th} \text{ obs.}$

= $3(\frac{8+1}{4} )^{th} \text{ obs.}$

= $(6.75 )^{th} \text{ obs.}$

$Q_3 = 6^{th} \text{ obs.} + 0.75[ 7^{th} \text{ obs.} -6^{th} \text{ obs.} ]$

= 4 + 0.75[6 - 4] = 5.5

So, the inter-quartile range of Team X's score is (5.5 - 0) = 5.5.

<u>Now, we will calculate the mean, median, range and inter-quartile range for Team Y;</u>

Mean of Team Y data is given by the following formula;

Mean, $\bar Y$ = $\frac{\sum Y}{n}$

= $\frac{4+ 2+ 0+ 3+ 2+ 1+ 6+ 4}{8}$ = $\frac{22}{8}$ = 2.75

So, the mean of Team Y's scores is 2.75.

Now, for calculating the median; we have to arrange the data in ascending order and then observe that the number of observations (n) in the data is even or odd.

Team Y: 0, 1, 2, 2, 3, 4, 4, 6

If n is odd, then the formula for calculating median is given by;

Median = $(\frac{n+1}{2} )^{th} \text{ obs.}$

If n is even, then the formula for calculating median is given by;

Median = $\frac{(\frac{n}{2})^{th} \text{ obs.} +(\frac{n}{2}+1)^{th} \text{ obs.} }{2}$

Here, the number of observations is even, i.e. n = 8.

So, Median = $\frac{(\frac{n}{2})^{th} \text{ obs.} +(\frac{n}{2}+1)^{th} \text{ obs.} }{2}$

= $\frac{(\frac{8}{2})^{th} \text{ obs.} +(\frac{8}{2}+1)^{th} \text{ obs.} }{2}$

= $\frac{(4)^{th} \text{ obs.} +(5)^{th} \text{ obs.} }{2}$

= $\frac{2+3}{2}$ = 2.5

So, the median of Team Y's score is 2.5.

Now, the range is calculated as the difference between the highest and the lowest value in our data.

Range = Highest value - Lowest value

= 6 - 0 = 6

So, the range of Team Y's score is 6.

Now, the inter-quartile range of the data is given by;

Inter-quartile range = $Q_3-Q_1$

$Q_1=(\frac{n+1}{4} )^{th} \text{ obs.}$

= $(\frac{8+1}{4} )^{th} \text{ obs.}$

= $(2.25 )^{th} \text{ obs.}$

$Q_1 = 2^{nd} \text{ obs.} + 0.25[ 3^{rd} \text{ obs.} -2^{nd} \text{ obs.} ]$

= 1 + 0.25[2 - 1] = 1.25

$Q_3=3(\frac{n+1}{4} )^{th} \text{ obs.}$

= $3(\frac{8+1}{4} )^{th} \text{ obs.}$

= $(6.75 )^{th} \text{ obs.}$

$Q_3 = 6^{th} \text{ obs.} + 0.75[ 7^{th} \text{ obs.} -6^{th} \text{ obs.} ]$

= 4 + 0.75[4 - 4] = 4

So, the inter-quartile range of Team Y's score is (4 - 1.25) = 2.75.

Hence, the correct statement is:

C. Team Y’s scores have a lower mean value.

4 0

3 years ago

A Company increased the number of its employees from 340 to 425. Find the percent increase.

Inessa [10]

25%

Because You must do the original number by the new.

so, 425-340= 85

but ur not done now,

divide that 85 by the original number then times by 100!

85/340= .25*100= 25%

and ur done hope this helped

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2 years ago

I need help with how to reason it

artcher [175]

Answer:

X=80

Step-by-step explanation:

As 100⁰ is Opposite to B and AD and EH Are PARALLEL so B=100⁰

180-100

=80⁰

<h2>MARK ME BRAINLIEST </h2>

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A survey taken during class revealed 30% of the students in a class bring their lunch to school. If the probability that a rando

ohaa [14]

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Daniel works at a nearby electronics store. He makes a commission of 15% on everything he sells. If he sells a laptop for $293.0

nata0808 [166]

$43.95 (293.00-15%=$43.95)

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2 years ago

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