Mathematics, 17.03.2020 20:09 HighSchool97654

At a rally attended by 520 people, 30% were fantes, 25% were ewes, 15% were nzemas, 20% were gas and the rest were gonjas. How many gonjas were at thzebedial e rally? How many more fantes than nzemas were at the rally? Draw a pie chart to illustrate the information ,

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Answer from: livewithgod

There were 52 gonjas in thzebedial rally.

There were 78 more fantes than nzemas.

Attached is the pie diagram

Step-by-step explanation:

Given,

Total number of people at the rally = 520

we need to find the number of gonjas at the rally.

Solution,

Firstly we will find out the number of fantes, ewes, nzemas and gas.

According to question, 30% are fantes.

So 30% of total number of people are fantes.

So we can frame it as;

Number of fantes = $30\%\times520$

Here we have to convert the percentile,

Number of people who were fantes = $\frac{30}{100}\times 520 = 156$

Also Given:

25% were ewes.

Number of people who were ewes = $\frac{25}{100}\times 520 = 130$

Now Given:

15% were nzemas

Number of people who were nzemas = $\frac{15}{100}\times 520 = 78$

Also Given:

20% were gas

Number of people who were gas = $\frac{20}{100}\times 520 =104$

Also Given:

the rest of the people were Gonjas

Number of people who were Gonjas = Total number of people attended the rally - Number of people who were ewes - Number of people who were nzemas - Number of people who were fantes - Number of people who were gas.

Number of people who were Gonjas = 520-156-130-78-104 = 52

Hence there were 52 gonjas in thzebedial rally.

Now we will find how many more fantes than nzemas were at the rally.

To find how many more fantes than nzemas were at the rally we will subtract Number of people who were nzemas from Number of people who were fantes.

framing in equation form we get;

more fantes than nzemas = 156-78=78

Hence there were 78 more fantes than nzemas.

At a rally attended by 520 people, 30% were fantes, 25% were ewes, 15% were nzemas, 20% were gas and

Answer from: Quest

answer: the answer is (b) amie should have multiplied

$54~\textup{m}^3~\textup{by}~\dfrac{4}{3}\pi.$

step-by-step explanation: given a sphere and a cylinder with same radius and height.

let, 'r' be the radius of the cylinder and sphere and 'h' be the height of the cylinder. according to the question,

r = h.

now, the volume of the sphere will be

$v_s=\pi r^2h=\pi r^3,$

and the volume of the cylinder will be

$v_c=\dfrac{4}{3}\pi r^3=54~\textup{m}^3.$

therefore, we have

$v_c=\dfrac{4}{3}v_s=\dfrac{4}{3}\times 54=72~\textup{m}^3.$

thus, ammie should have multiplied the volume of the sphere $54~\textup{m}^3~\textup{by}~\dfrac{4}{3}\pi.$

Answer from: Quest

i believe that the answer is

b. y-1=-2(x-4)

hope this !

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