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3 years ago

There are 25 students in the

Mathematics

1 answer:

4vir4ik [10]3 years ago

4 0

Answer:

ratio 1- 23:25

and the rest i dont know sorry. LOL

Step-by-step explanation:

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twenty students in A and 20 students in class B were asked how many hours they took to prepare for an exam.The data sets represe

SSSSS [86.1K]

Answer:

is there a graph i can read?

Step-by-step explanation:

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2 years ago

The quotient of a number and 15 is no greater than 450. What are the possible values. For the number?

FinnZ [79.3K]

A number can be represented by n. So n/15≤450 then multiply both sides by 15 to get n by itself n≤6,750 Hope this helped!!! :)

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3 years ago

Using the integer tiles interactive, what steps do you need to do to evaluate the following expression? Select two options. –5(–

dangina [55]

The steps you need to do to evaluate the expression are (b) Drag zero pairs to the window. and (c) Remove the 5 groups of –2 tiles.

<h3>How to determine the steps you need to do to evaluate the expression</h3>

The expression is given as:

-5(-2)

The above expression represents the product of -5 and -2

This can be interpreted as:

Removal of 5 groups of -2; to remove means negative
Dragging zero pairs of 5 and 2 to the interactive window

Hence, the steps you need to do to evaluate the expression are (b) Drag zero pairs to the window. and (c) Remove the 5 groups of –2 tiles.

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8 0

1 year ago

How many terms are there in the sequence 1, 8, 28, 56, ..., 1 ?

BabaBlast [244]

Answer:

9 terms

Step-by-step explanation:

Given:

1, 8, 28, 56, ..., 1

Required

Determine the number of sequence

To determine the number of sequence, we need to understand how the sequence are generated

The sequence are generated using

$\left[\begin{array}{c}n&&r\end{array}\right] = \frac{n!}{(n-r)!r!}$

Where n = 8 and r = 0,1....8

When r = 0

$\left[\begin{array}{c}8&&0\end{array}\right] = \frac{8!}{(8-0)!0!} = \frac{8!}{8!0!} = 1$

When r = 1

$\left[\begin{array}{c}8&&1\end{array}\right] = \frac{8!}{(8-1)!1!} = \frac{8!}{7!1!} = \frac{8 * 7!}{7! * 1} = \frac{8}{1} = 8$

When r = 2

$\left[\begin{array}{c}8&&2\end{array}\right] = \frac{8!}{(8-2)!2!} = \frac{8!}{6!2!} = \frac{8 * 7 * 6!}{6! * 2 *1} = \frac{8 * 7}{2 *1} =2 8$

When r = 3

$\left[\begin{array}{c}8&&3\end{array}\right] = \frac{8!}{(8-3)!3!} = \frac{8!}{5!3!} = \frac{8 * 7 * 6 * 5!}{5! *3* 2 *1} = \frac{8 * 7 * 6}{3 *2 *1} = 56$

When r = 4

$\left[\begin{array}{c}8&&4\end{array}\right] = \frac{8!}{(8-4)!4!} = \frac{8!}{4!3!} = \frac{8 * 7 * 6 * 5 * 4!}{4! *4*3* 2 *1} = \frac{8 * 7 * 6*5}{4*3 *2 *1} = 70$

When r = 5

$\left[\begin{array}{c}8&&5\end{array}\right] = \frac{8!}{(8-5)!5!} = \frac{8!}{5!3!} = \frac{8 * 7 * 6 * 5!}{5! *3* 2 *1} = \frac{8 * 7 * 6}{3 *2 *1} = 56$