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AlladinOne [14]
2 years ago
15

A game room has a floor that is 200 feet by 40 feet. A scale drawing of the floor on grid paper uses a scale of 2 units:5 feet.

What are the dimensions of the scale drawing
Mathematics
1 answer:
LiRa [457]2 years ago
4 0
200 FEET
2 units/5 feet = x units/200 feet
x = 2/5 (200) = 400/5 = 80 units

Cross Multiplication
5x = 2(200)
x = 400/5 = 80 units

40 FEET
2 units/5 feet = x units/40 feet
x = 2/5(40) = 80/5 = 16

Cross Multiply
2/5 = x/40
5x=80
x = 80/5 = 16

Dimensions of Scale Drawing:
80units × 16units
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On a coordinate plane, a line goes through (negative 4, 0) and (4, negative 4). A point is at (2, 3).
Valentin [98]

9514 1404 393

Answer:

  (b)  x +2y = 8

Step-by-step explanation:

The only offered line that includes the given point is ...

  x +2y = 8

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We can check the other choices:

  x + 2y = 2 +2(3) = 2+6 = 8 . . . matches B (not A)

  2x +y = 2(2) +3 = 4+3 = 7 . . . . not a choice

_____

<em>Getting there from scratch</em>

The standard form equation for a line can be written from ...

  (y2 -y1)x -(x2 -x1)y = constant

  (-4 -0)x -(4 -(-4))y = constant . . . . using the given points (-4, 0) and (4, -4)

  -4x -8y = constant

For standard form, we need the leading coefficient to be positive, and we need common factors removed. We can get there by dividing by -4.

  x +2y = constant

The value of the constant will be whatever it takes for the given point to lie on the line. For (x, y) = (2, 3) to be a solution, we must have ...

 x +2y = (2) +2(3) = constant = 8

The desired line has the equation ...

  x +2y = 8

8 0
2 years ago
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How does the experimental probability of rolling a 3 compare with the theoretical probability of rolling a 3​
Ne4ueva [31]

Answer:The experimental probability of rolling a 3 is 1/30 greater than the theoretical probability of rolling a 3.

Step-by-step explanation:i did the test and got it right

6 0
2 years ago
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An example problem in a Statistics textbook asked to find the probability of dying when making a skydiving jump.
MArishka [77]

Answer:

(a) 0.999664

(b) 15052

Step-by-step explanation:

From the given data of recent years,  there were about 3,000,000 skydiving jumps and 21 of them resulted in deaths.

So, the probability of death is \frac{21}{3000000}==0.000007.

Assuming, this probability holds true for each skydiving and does not change in the present time.

So, as every skydiving is an independent event having a fixed probability of dying and there are only two possibilities, the diver will either die or survive, so, all skydiving can be regarded as is Bernoulli's trial.

Denoting the probability of dying in a single jump by q.

q=7\times 10^{-6}=0.000007.

So, the probability of survive, p=1-q

\Rightarrow p=1-7\times 10^{-6}=0.999993.

(a) The total number of jump he made, n=48

Using Bernoulli's equation, the probability of surviving in exactly 48 jumps (r=48) out of 48 jumps (n=48) is

=\binom(n,r)p^rq^{n-r}

=\binom(48,48)(0.999993)^{48}(0.000007)^{48-48}

=(0.999993)^{48}=0.999664 (approx)

So, the probability of survive in 48 skydiving is 0.999664,

(b) The given probability of surviving =90%=0.9

Let, total n skydiving jumps required to meet the surviving probability of 0.9.

So, By using Bernoulli's equation,

0.9=\binom {n }{r} p^rq^{n-r}

Here, r=n.

\Rightarrow 0.9=\binom{n}{n}p^nq^{n-n}

\Rightarrow 0.9=p^n

\Rightarrow 0.9=(0.999993)^n

\Rightarrow \ln(0.9)=n\ln(0.999993) [ taking \log_e both sides]

\Rightarrow n=\frac {\ln(0.9)}{\ln(0.999993)}

\Rightarrow n=15051.45

The number of diving cant be a fractional value, so bound it to the upper integral value.

Hence, the total number of skydiving required to meet the 90% probability of surviving is 15052.

3 0
3 years ago
Let Y1 and Y2 have the joint probability density function given by:
Ann [662]

Answer:

a) k=6

b) P(Y1 ≤ 3/4, Y2 ≥ 1/2) =  9/16

Step-by-step explanation:

a) if

f (y1, y2) = k(1 − y2), 0 ≤ y1 ≤ y2 ≤ 1,  0, elsewhere

for f to be a probability density function , has to comply with the requirement that the sum of the probability of all the posible states is 1 , then

P(all possible values) = ∫∫f (y1, y2) dy1*dy2 = 1

then integrated between

y1 ≤ y2 ≤ 1 and 0 ≤ y1 ≤ 1

∫∫f (y1, y2) dy1*dy2 =  ∫∫k(1 − y2) dy1*dy2 = k  ∫ [(1-1²/2)- (y1-y1²/2)] dy1 = k  ∫ (1/2-y1+y1²/2) dy1) = k[ (1/2* 1 - 1²/2 +1/2*1³/3)-  (1/2* 0 - 0²/2 +1/2*0³/3)] = k*(1/6)

then

k/6 = 1 → k=6

b)

P(Y1 ≤ 3/4, Y2 ≥ 1/2) = P (0 ≤Y1 ≤ 3/4, 1/2 ≤Y2 ≤ 1) = p

then

p = ∫∫f (y1, y2) dy1*dy2 = 6*∫∫(1 − y2) dy1*dy2 = 6*∫(1 − y2) *dy2 ∫dy1 =

6*[(1-1²/2)-((1/2) - (1/2)²/2)]*[3/4-0] = 6*(1/8)*(3/4)=  9/16

therefore

P(Y1 ≤ 3/4, Y2 ≥ 1/2) =  9/16

8 0
3 years ago
The point (4,-2) is the vertex of the graph of a quadratic function. The points (8,6) and (2,0) also fall on the graph of the fu
Elina [12.6K]

Answer:

Step-by-step explanation:

y = ax² + bx + c

~~~~~~~

(4, - 2), (8, 6), (2, 0)

a(4²) + b(4) + c = - 2

a(8²) + b(8) + c = 6

a(2²) + b(2) + c = 0

16a + 4b + c = - 2 .............. <em>(1)</em>

64a + 8b + c = 6 ............... <em>(2)</em>

4a + 2b + c = 0 ................. <em>(3)</em>

a = 0.5 ; b = - 4 ; c = 6

<em>y = 0.5x² - 4x + 6</em>

Part A: (0, 6), (6, 0)

Part B: (0,6)

Part C: (6, 0)

Part D: [ 4, ∞ )

4 0
3 years ago
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