What algebraic property is demonstrated in the equation below? 5(8× - 4) - 10 = 40x - 20

In 1 and 2, list multiples of each number to find the LCM of each pair of number.

Tanya [424]

#1

Multiples of 2:

2, 4, 6, 8, 10, 12, 14, 16, 18, 20,...

Multiples of 5:

5, 10, 15, 20, 25, 30, 35, 40, ...

#2

Multiples of 6:

6, 12, 18, 24, 30, 36, 42, 48, 54, 60, ...

Multiples of 10:

10, 20, 30, 40, 50, 60, ...

6 0

2 years ago

51:30

Softa [21]

Answer:

The correct option is 'c' cross-multiplication

Step-by-step explanation:

5 0

1 year ago

The SAT scores have an average of 1200 with a standard deviation of 60. A sample of 36 scores is selected. a) What is the probab

LiRa [457]

Answer:

a) 0.0082

b) 0.9987

c) 0.9192

d) 0.5000

e) 1

Step-by-step explanation:

The question is concerned with the mean of a sample.

From the central limit theorem we have the formula:

$z=\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$

a) $z=\frac{1224-1200}{\frac{60}{\sqrt{36} } }=2.40$

The area to the left of z=2.40 is 0.9918

The area to the right of z=2.40 is 1-0.9918=0.0082

$\therefore P(\bar X\:>\:1224)=0.0082$

b) $z=\frac{1230-1200}{\frac{60}{\sqrt{36} } }=3.00$

The area to the left of z=3.00 is 0.9987

$\therefore P(\bar X\:$

c) The z-value of 1200 is 0

The area to the left of 0 is 0.5

$z=\frac{1214-1200}{\frac{60}{\sqrt{36} } }=1.40$

The area to the left of z=1.40 is 0.9192

The probability that the sample mean is between 1200 and 1214 is

$P(1200\:$

d) From c) the probability that the sample mean will be greater than 1200 is 1-0.5000=0.5000

e) $z=\frac{73.46-1200}{\frac{60}{\sqrt{36} } }=-112.65$

The area to the left of z=-112.65 is 0.

The area to the right of z=-112.65 is 1-0=1

5 0

2 years ago

A group of randomly selected members of The Foodies Food Club were asked to pick their favorite type of food. The bar graph belo

zloy xaker [14]

Answer: The answer is B

3 0

3 years ago

Read 2 more answers

Please answer the following questions

laila [671]

Answer:

4a) 110 square centimetres

4b) 127 square centimetres

6) 292 square centimetres

8) 800 tiles

Step-by-step explanation:

4. We need to find the area of the large rectangle and then deduct the area of the unshaded part:

a) The large rectangle has dimensions 12 cm by 15 cm. Its area is:

A = 12 * 15 = 180 square centimetres

The unshaded part has a length of 15 - (3 + 2) cm i.e. 10 cm and a width of 7 cm. Its area is:

a = 10 * 7 = 70 square centimetres

Therefore, the area of the shaded part is:

A - a = 180 - 70 = 110 square centimetres

b) The large rectangle has dimensions 13 cm by 11 cm. Its area is:

A = 13 * 11 = 143 square centimetres

The unshaded part has dimensions 8 cm by 2 cm. Its area is:

a = 8 * 2 = 16 square centimetres

Therefore, the area of the shaded part is:

A - a = 143 - 16 = 127 square centimetres

6. The background area of the space not covered by the photograph is the area of the frame minus the area of the photograph.

The frame has dimensions 24 cm by 18 cm. Therefore, its area is:

A = 24 * 18 = 432 square centimetres

The photograph has dimensions 14 cm by 10 cm. Therefore, its area is:

a = 14 * 10 = 140 square centimetres

Therefore, the background area of the space not covered by the photograph is:

A - a = 432 - 140 = 292 square centimetres

8) The floor has dimensions 8 m by 4 m. The area of the floor is:

A = 8 * 4 = 32 square centimetres

Each square tile has dimensions 20 cm by 20 cm. In metres, that is 0.2 m by 0.2 m. The area of each tile is:

a = 0.2 * 0.2 = 0.04 square metres

The number of tiles that are needed is the area of the floor divided by the area of each tile:

A / a = 32 / 0.04 = 800 tiles

4 0

3 years ago

What algebraic property is demonstrated in the equation below? 5(8× - 4) - 10 = 40x - 20 - 10