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2 years ago

Solve the proportion x-9/3 = 5/13

Mathematics

1 answer:

r-ruslan [8.4K]2 years ago

7 0

13x-117=15
13x=15+117
13x=132
x=132/13
x=10.1538

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Can somebody plz help answer all these questions (only if u know how to do it) lol thanks!!!

vladimir1956 [14]

Answer:

13=6 14=5. 15=3 16=2 17=5 18= 7 19=2 20=9

7 0

2 years ago

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A sample of men's heights was taken and the mean was 68.8 inches. The standard deviation is 2.8 inches. What percent of the men

jarptica [38.1K]

Answer:

87.29%

Step-by-step explanation:

Given: Mean= 68.8 inches

Standard deviation= 2.8 inches

Now, finding the percent of the men in the sample were greater than 72 inches.

We know, z-score= $\frac{x-mean}{standard\ deviation}$

z-score= $\frac{72-68.8}{2.8}$

⇒ z-score= $\frac{3.2}{2.8} = 1.14$

∴ z-score= 1.14

Next, using normal distribution table to find percentage.

Coverting 0.8729 into percentage= $0.8729\times 100$

We get the percentage as 87.29%

Hence, 87.29% of the men in the sample were greater than 72 inches.

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2 years ago

1/x + 1/3x =6 how do you solve this equation for x

Elodia [21]

Answer:

x = 2/9 (or = 0.222)

Step-by-step explanation:

1/x + 1/(3x) =6 (multiply both sides by 3x)

(1/x) (3x) + [1/(3x)](3x) =6 (3x)

3 + 1 = 18x

4 = 18x

x = 4/18

x = 2/9 (or = 0.222)

3 0

2 years ago

BRAINLIEST !!! Can someone add me as a friend plz need to look someone up but cant till someone adds me

eimsori [14]

Answer:

I have done what you want

4 0

2 years ago

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Suppose that a spherical droplet of liquid evaporates at a rate that is proportional to its surface area: where V = volume (mm3

Alex

Answer:

V = 20.2969 mm^3 @ t = 10

r = 1.692 mm @ t = 10

Step-by-step explanation:

The solution to the first order ordinary differential equation:

$\frac{dV}{dt} = -kA$

Using Euler's method

$\frac{dVi}{dt} = -k *4pi*r^2_{i} = -k *4pi*(\frac {3 V_{i} }{4pi})^(2/3)\\ V_{i+1} = V'_{i} *h + V_{i} \\$

Where initial droplet volume is:

$V(0) = \frac{4pi}{3} * r(0)^3 = \frac{4pi}{3} * 2.5^3 = 65.45 mm^3$

Hence, the iterative solution will be as next:

i = 1, ti = 0, Vi = 65.45

$V'_{i} = -k *4pi*(\frac{3*65.45}{4pi})^(2/3) = -6.283\\V_{i+1} = 65.45-6.283*0.25 = 63.88$

i = 2, ti = 0.5, Vi = 63.88

$V'_{i} = -k *4pi*(\frac{3*63.88}{4pi})^(2/3) = -6.182\\V_{i+1} = 63.88-6.182*0.25 = 62.33$

i = 3, ti = 1, Vi = 62.33

$V'_{i} = -k *4pi*(\frac{3*62.33}{4pi})^(2/3) = -6.082\\V_{i+1} = 62.33-6.082*0.25 = 60.813$

We compute the next iterations in MATLAB (see attachment)

Volume @ t = 10 is = 20.2969

The droplet radius at t=10 mins

$r(10) = (\frac{3*20.2969}{4pi})^(2/3) = 1.692 mm\\$

The average change of droplet radius with time is:

Δr/Δt = $\frac{r(10) - r(0)}{10-0} = \frac{1.692 - 2.5}{10} = -0.0808 mm/min$

The value of the evaporation rate is close the value of k = 0.08 mm/min

Hence, the results are accurate and consistent!

5 0

2 years ago