there are 102 to teachers in a school of 3300 students find the ratio of the number of teachers to the number of student​

Answer:

Volume of removed water = Length X width X decrease in water level

⇒ 18 = 20 X width X 0.15

∴ width = 6 metre

Step-by-step explanation:

Given:

A tank is in the form of a cuboid whose length is 20 metres. If 18 cubic metres of water is removed from it the water level goes down by 15 centimetres.

To find:

The width of the tank.

Solution:

The width of the tank is 6m.

To answer this question, we will follow the following steps:

As given, we have,

The length of a tank which is in the form of cuboid = 20m

Now,

When 18 cubic metres of water is removed, the water level goes down by = 15cm = 0.15m (as 1m = 100cm)

Now,

We know that the volume of a cuboid is equal to the product of length, breadth and height of that cuboid.

Here,

The volume of removed water

= length × width × decreased height of water

So,

[tex]20 \times width \times 0.15 = 18[/tex]

[tex]3 \times width = 18[/tex]

[tex]width = 6m[/tex]

Hence, the width of the tank is 6m.

Step-by-step explanation:

इसमें क्या बताना है हां बताओ क्या बताना

Answer:

1.Earth orbits the Sun at an average distance of 149.60 million km (92.96 million mi) in a counterclockwise pattern viewed above the northern hemisphere. One complete orbit takes 365.256 days (1 sidereal year), during which time Earth has traveled 940 million km (584 million mi).

2.Let m and M be the masses of the earth and the sun respectively and v the speed of the earth in circular orbit. To keep the earth in circular orbit, the gravitational force.

reaming answer is the photo

Step-by-step explanation:

please mark me as brainlist

hope it's helpful

Answer:

The state is divided into 9 divisions, 38 districts, 101 subdivisions, and 534 circles. 12 municipal corporations, 49 Nagar Parishads, and 80 Nagar Panchayats for administrative purposes.

Division Headquarters Districts District map of Bihar

Patna Patna Bhojpur, Buxar, Kaimur, Patna, Rohtas, Nalanda Divisions of Bihar.svg

Saran Chapra Saran, Siwan , Gopalganj

Tirhut Muzaffarpur East Champaran, Muzaffarpur, Sheohar, Sitamarhi, Vaishali, West Champaran

Purnia Purnia Araria, Katihar, Kishanganj, Purnia

Bhagalpur Bhagalpur Banka, Bhagalpur

Darbhanga Darbhanga Darbhanga, Madhubani, Samastipur

Kosi Saharsa Madhepura, Saharsa, Supaul

Magadh Gaya Arwal, Aurangabad, Gaya, Jehanabad, Nawada

Munger Munger Begusarai,Jamui, Khagaria, Munger, Lakhisarai, Sheikhpura

Explanation:

hope this helps

See AP (Arithmetic Progression)

a+(n-1)d=1+(71-1)1=71 [a=1,n=71,d=1]

Sum= =n[2a+(n-1)b]/2 =71[2+(71-1)1]/2=71[2+70]/2=(71×72)/2=5112/2=2556.

Average = sum/ no. of observations [here, no. of natural nos.]=2556/71=36.

Answer:

The answer is 36

Given problem:

find the average of first 71 natural numbers ? ​

Step-by-step explanation:

We know the first 71 natural number series is  1, 2, 3, 4... 71  

Here 1, 2, 3, 4.. 71 is in the form of a AP

where first term  [tex]a[/tex] = 1,  last term [tex]l[/tex] = 71  and number of terms = 71  

therefore, sum of terms = [tex]\frac{n(a+l)}{2}[/tex]  = [tex]\frac{71(1+71)}{2}[/tex] = [tex]\frac{71(72)}{2}[/tex] = 71(36) = 2556

Average of first 71 natural numbers

         = sum of 71 numbers / 71

         = 2556 / 71 = 36