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4 years ago

Find x (7x-11) (2x-3) (5x-2)

Mathematics

1 answer:

luda_lava [24]4 years ago

6 0

Answer:

70x -16

Step-by-step explanation:

(7x-11) (2x-3) (5x-2)

7x*2x*5x * -11*-3*-2

70x -16

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If the measures of 2 angles of a triangle are 35 degree and 80 degree then find the measure of its third angle

likoan [24]

Step-by-step explanation:

the 3rd angle = 180 - ( 35+80 ) = 65

7 0

3 years ago

Read 2 more answers

Help me with these simple math problems

earnstyle [38]

Answer:

Please read the answer below.

Step-by-step explanation:

1. Australia:

75 * 1.87 =140 Australian dollars

2. Brazil:

75 * 2.32 = 174 Reals

3. Britain:

75 * 0.69 = 52 Pounds

4. Canada:

75 * 1.60 = 120 Canadian dollars

5. China:

75 * 8.28 = 621 Yuan

6. Denmark:

75 * 8.43 = 632 Kroner

7. Japan:

75 * 131.55 = 9,866 Yen

8. Mexico:

75 * 9.19 = 689 Mexican pesos

9. South Africa:

75 * 11.9 = 893 Rands

10. Sweden:

75 * 10.61 = 796 Kronor

11. Switzerland:

75 * 1.68 = 126 Francs

12. Thailand:

75 * 44.18 = 3,314 Baht

All currencies rounded to the next integer.

Note: Same answer to question 14454918, answered by me today.

8 0

4 years ago

Compare and contrast division property of equality and inequality

ch4aika [34]

The two properties are exactly the same when dividing by a positive number. For the division property of equality, dividing by a negative number causes the equal sign to change. Dividing both sides of an inequality by a negative number does not have an effect on the relation symbol.

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4 years ago

Simplify the expression fully <br> (3x2 + 7y) (2x2 – 3x + 9)

allsm [11]

(3x2 + 7y) (2x2 – 3x + 9)
=(6+7y)(-3x+13)
=-18x+78-21xy+91y

3 0

3 years ago

Suppose that 7 in every 10 auto accidents involve a single vehicle. If 15 auto accidents are randomly selected, compute the prob

kobusy [5.1K]

Answer:

$\mathbf{P(X \le 4 ) \simeq 0.0006722}$

Step-by-step explanation:

From the information given:

p = x/n

p = 7/10

p = 0.7

sample size n = 15

Suppose X be the number of accidents involved by a single-vehicle.

Then;

$X \sim Binom (15,0.7)$

Thus, the required probability that at most 4 involve in a single-vehicle is

$P(X\le 4) \\ \\P(X \le 4) = P(X = 0) + P(X =1 ) + ... + P(X = 4)$

$P(X \le 4 ) = (^{15}_0) *0.7^0 *0.3^{15-0} + (^{15}_1) *0.7^1 *0.3^{15-1} + (^{15}_2) *0.7^2 *0.3^{15-2} + (^{15}_3) *0.7^3 *0.3^{15-3} + (^{15}_4) *0.7^4 *0.3^{15-4}$

$P(X \le 4 ) = (\dfrac{15!}{0!(15-0)!}) *0.7^0 *0.3^{15-0} + (\dfrac{15!}{1!(15-1)!}) *0.7^1 *0.3^{15-1} + (\dfrac{15!}{2!(15-2)!}) *0.7^2 *0.3^{15-2} + (\dfrac{15!}{3!(15-3)!}) *0.7^3 *0.3^{15-3} + (\dfrac{15!}{4!(15-4)!}) *0.7^4 *0.3^{15-4}$

$P(X \le 4 ) =1.4348907 \times 10^{-8} +5.02211745 \times 10^{-7} + 8.20279183 \times 10^{-6} + 8.29393397 \times 10^{-5} + 5.80575378 \times 10^{-4}$

$P(X \le 4 ) =6.7223407 \times 10^{-4}$

$\mathbf{P(X \le 4 ) \simeq 0.0006722}$

3 0

3 years ago