১.১ স্প্রিং তুলার সাহায্যে যে রাশি মাপা হয় তা হলাে- (ক) ঘনত্ব (খ) আয়তন (গ) ওজন (ঘ) ত্বরণ ​

Answer:

c) ⏏ ⏏ ⏏ drive

Explanation:

• The answer is in the drive, which is often a small hole in the side of the computer. Sometimes the drive is at the front, and sometimes at the back. A drive is usually the only physical thing going from the computer to a disk or CD.
• The drive is often the only physical thing that goes from the computer to the disk or CD.
• The drive does not matter much, though. So many computers have drives that do not matter—all computers have a disk drive, and some computers have a CD drive, too, although CD drives are less common.
• The drive is the one physical part of the computer that will cause you problems if you don't know how to replace it.

Answer:

dear raj the package that we were waiting that dress do u remember are designed

Explanation:

Explanation:

Classes :-

5-7 , 7-9 , 9-11 , 11-13 , 13-15 , 15-17

Frequency :-

55, 55, 70, 150, 86, 84

Cumulative frequencies :-

55, 110, 180, 330, 416 , 500

Sum of all frequencies (N)

= 55+55+70+150+86+84

= 500

Therefore, N = 500

N/2 = 500/2 = 250

250 lies in the cumulative frequency 330

Frequency of Median class = 150

Class interval of Median class = 11-13

Lower boundary of the median class (l) = 11

[tex]\begin{gathered}\boxed{\begin{array}{c|c|c} \bf Classes & \bf Frequency & \bf Cumulative\; Frequency \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 5 - 7 & 55& 55 \\ \\ \sf 7 - 9& 55 & 110\\ \\ \sf \: 9 - 11& \sf 70 & \sf 180\\ \\ 11 - 13 & 150 & 330 \\ \\ \sf 13 - 15 & \sf 86 & 416\\ \\ \sf \: 15 - 17& \sf \: 84 & \sf 500 \end{array}} \\ \end{gathered}[/tex]

To find the median class:

• ➡ First add the total number of frequencies.

[tex]\rm \implies \: 55 +55+ 70 + 150 + 86+84 [/tex]

[tex]\bf\implies \: 500[/tex]

Here , we know that

• ➡ f = 500

Now ,

[tex]\rm \implies \: \dfrac{(Total \: number \: of \: frequency)}{2} [/tex]

• Put the given values in this and solve

[tex]\rm \implies \dfrac{500}{2} = 250[/tex]

Henceforth,

• ➡ The value 250 lies in the class interval 11-13 from the cummulative frequency table.

• ➡ As 330 is just greater than 250,

[tex]\rm \longrightarrow \: \therefore \: Median \: class = 11 - 13[/tex]

[tex] \rm \therefore \: The \: lower \: limit \: of \: the \: modal \: class \: is \: 11.[/tex]

Answer:

a. penna

Explanation:

the other river are in india

(a) Brahmaputra

The river Brahmaputra is located in the eastern mountain ranges of India while the rest of the mentioned rivers flow in the western and southern plateau regions.

At an elevation of 5300 metres, the river begins in the Himalayan Kailash mountains. It enters India through Arunachal Pradesh and then flows through Assam and Bangladesh before joining the Bay of Bengal.