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3 years ago

What are equivalent to 2:9

Mathematics

1 answer:

dimaraw [331]3 years ago

6 0

Answer:

$4:18$

Step-by-step explanation:

<u>Step 1: Multiply both sides by 2 </u>

$2:9$

$(2*2) : (2*9)$

$4:18$

Answer: $4:18$

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Suppose that 40 percent of the drivers stopped at State Police checkpoints in Storrs on Spring Weekend show evidence of driving

lesantik [10]

Answer:

a) 0.778

b) 0.9222

c) 0.6826

d) 0.3174

e) 2 drivers

Step-by-step explanation:

Given:

Sample size, n = 5

P = 40% = 0.4

a) Probability that none of the drivers shows evidence of intoxication.

$P(x=0) = ^nC_x P^x (1-P)^n^-^x$

$P(x=0) = ^5C_0 (0.4)^0 (1-0.4)^5^-^0$

$P(x=0) = ^5C_0 (0.4)^0 (0.60)^5$

$P(x=0) = 0.778$

b) Probability that at least one of the drivers shows evidence of intoxication would be:

P(X ≥ 1) = 1 - P(X < 1)

$= 1 - P(X = 0)$

$= 1 - ^5C_0 (0.4)^0 * (0.6)^5$

$= 1 - 0.0778$

$= 0.9222$

c) The probability that at most two of the drivers show evidence of intoxication.

P(x≤2) = P(X = 0) + P(X = 1) + P(X = 2)

$^5C_0 (0.4)^0 (0.6)^5 + ^5C_1 (0.4)^1 (0.6)^4 + ^5C_2 (0.4)^2 (0.6)^3$

$= 0.6826$

d) Probability that more than two of the drivers show evidence of intoxication.

P(x>2) = 1 - P(X ≤ 2)

$= 1 - [^5C_0 (0.4)^0 (0.6)^5 + ^5C_1 (0.4)^1 (0.6)^4 + ^5C_2 * (0.4)^2 (0.6)^3]$

$= 1 - 0.6826$

$= 0.3174$

e) Expected number of intoxicated drivers.

To find this, use:

Sample size multiplied by sample proportion

n * p

= 5 * 0.40

= 2

Expected number of intoxicated drivers would be 2

7 0

3 years ago

What is the area of Shaded part?

svetlana [45]

Answer:

1π

Step-by-step explanation:

suppose the radius of semicircle P is r,

then the radius of semicircle Q = (r+d)/2 ... d≤r

radius of semicircle R = (r-d)/2

area P = 1/2 (r)²π

area Q = 1/2 ((r+d)/2)² π = 1/8 (r² + 2rd + d²)π

area R = 1/2 ((r-d)/2)² π = 1/8 (r² - 2rd + d²)π

shaded area = P-Q-R = 1/2 r²π - 1/4 (r² + d²)π

= ((r² - d²)/4) * π

because there is no constant r value in the question and d value changes with the r change, when the vertical segment length equal the semicircle P radius (r), r=2 and d = 0

therefore the shaded area = ((2² -0²)/4)*π = 1π

8 0

2 years ago

Twice a number added to a smaller number is 5. The difference of 5 times the smaller number and the larger number is 3. Let x re

Sergio [31]

Answer:

1) 2Y+X=5

2) 5X-Y=3

6 0

3 years ago

WILL MARK BRAINLIEST

alexandr1967 [171]

Answer:

Equation: 7x - 23 = 25 + x

Solution: x = 8

Step-by-step explanation:

7x - 23 = 25 + x

1. Subtract 25 from both sides:

7x - 23 - 25 = 25 - 25 + x

7x - 48 = x

2. Subtract 7x from both sides to get x on one side:

7x - 7x - 48 = x - 7x

- 48 = -6x

3. Divide both sides by -6 to get x by itself:

8 = x

8 is the solution

4 0

3 years ago

Jerome burns 4 cal/min walking and 10 cal/min running. He walks between 10 and 20 min each day and runs between 30 and 45 min ea

erik [133]

Answer:

20 minutes should spend on walking and 30 minutes on running.

Step-by-step explanation:

Let x represents the minutes spent on walking and y represents the minutes spent on running,

∵ He burns 4 cal/min walking and 10 cal/min running,

So, the total calories burnt,

Z = 4x + 10y

Which has to be maximised.

Now, he walks between 10 and 20 min each day,

i.e. 10 < x < 20,

Also, he runs between 30 and 45 min each day,

i.e. 30 < y < 45,

By graphing 10 < x < 20 and 30 < y < 45,

We obtained the vertices of feasible region,

(10, 45), (20, 45), (10, 30) and (20, 30)

At (10, 45),

Z = 4(10) + 10(45) = 40 + 450 = 490,

At (20, 45),

Z = 4(20) + 10(45) = 80 + 450 = 530,

At (10, 30),

Z = 4(10) + 10(30) = 40 + 300 = 340,

At (20, 30)

Z = 4(20) + 10(30) = 80 + 300 = 380

Hence, maximum calories is 530 when 20 minutes spent on walking and 30 minutes spent on running.

5 0

3 years ago