When completely for a cube shaped container will hold 8000 cm³ of water. What is the length of an edge of the container

Which system of linear equations appears to have a solution of (3, 0)?

Alenkasestr [34]

Answer:

C

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

D

yarga [219]

Answer:

I think the answer is 67.5. Hope this helps! :)

3 0

3 years ago

Suppose that three computer boards in a production run of forty are defective. A sample of five is to be selected to be checked

IgorC [24]

Answer:

(a) 658,008 different samples can be chosen.

(b) 222,111 samples will contain at least one defective board.

(c) The probability that a randomly chosen sample of five contains at least one defective board is 0.34.

Step-by-step explanation:

We are given that three computer boards in a production run of forty are defective. A sample of five is to be selected to be checked for defects.

(a) To find how many different samples can be chosen, we will use a combination formula here because the order of selecting a sample of 5 from the production run of 40 doesn't matter.

Here, n = total sample = 40 and r = selected sample = 5

So, the combination formula is; $^{n}C_r= \frac{n!}{r! \times (n-r)!}$

$^{40}C_5= \frac{40!}{5! \times (40-5)!}$

$^{40}C_5= \frac{40!}{5! \times 35!}$

$^{40}C_5$ = 658,008 ways

So, 658,008 different samples can be chosen.

(b) To find how many samples will contain at least one defective board, we will first find how many samples will contain no or 0 defective board.

For this also, we will use a combination where n = 40 - 3 = 37 non-defective computer board and a sample of r = 5 computer boards.

So, $^{n}C_r= \frac{n!}{r! \times (n-r)!}$

$^{37}C_5= \frac{37!}{5! \times (37-5)!}$

$^{37}C_5= \frac{37!}{5! \times 32!}$

$^{37}C_5$ = 435,897 ways

This means that 435,897 of the 658,008 samples will contain no defective board.

Now, the samples that will contain at least one defective board = Total samples - Samples that contain no defective board

= 658,008- 435, 897

= 222,111

(c) The probability that a randomly chosen sample of five contains at least one defective board is given by;

Required Probability = $\frac{222,111}{658,008}$

= 0.34 or 34%

3 0

4 years ago

The length and breadth of a rectangle are in the ratio 3:2 respectively. If the sides of the rectangle are increased by 1 meter

Aleks04 [339]

ANSWER:

The length and breadth of the rectangle are 18 m and 12 m.

SOLUTION:

Let the length and breadth of a rectangle be "l" and "b"

Given,length and breadth of the rectangle are in ratio 3 : 2

Then, length : breadth :: 3 : 2

$\begin{array}{l}{\frac{l}{b}=\frac{3}{2}} \\\\ {l=\frac{3 b}{2}}\end{array}$ -- eqn 1

After changing the length and breadth by 1 meter on both sides, length and breadth becomes L+2 and b+2

Now, the ratio of length to breadth is 10 : 7

Length : breadth :: 10 : 7

$\frac{l+2}{b+2}=\frac{10}{7}$

$(l +2 ) \times 7 = 10 \times ( b + 2)$

7l + 14 = 10b + 20

10b – 7l + 20 -14 = 0

10b – 7l + 6 =0 -- eqn (2)

Now, substitute “l” value in (2)

$10 b-7\left(\frac{3 b}{2}\right)+6=0$

$10 b-\frac{21 b}{2}+6=0$

20b – 21b + 12 = 0

-b + 12 = 0

b = 12.

Substitute b value in (2)

10(12) – 7l + 6 = 0

120 + 6 = 7l

7l = 126

l = 18

hence, the length and breadth of the rectangle are 18 m and 12 m.

6 0

3 years ago

Use substitution to solve the following system of equations. y=3x +8 5x+ 2y = 5

Shkiper50 [21]

Answer:

5x+2(3x+8)

5x+6x+16

11x+16

11x=-16

X=16/11

6 0

3 years ago