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alexandr402 [8]
3 years ago
10

50 Points!

Mathematics
2 answers:
Lesechka [4]3 years ago
5 0

Answer:

D 1/4

Step-by-step explanation:

forsale [732]3 years ago
5 0

Answer:

D

Step-by-step explanation:

There are 8 sections, and since you are landing on an even number and then a yellow section, the probability will be out of 16. There are 4 even numbers and also 4 yellow sections, so it is going to be 4/16. You then reduce this fraction to get 1/4.

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give me a complicated question i give you a complicated answer now translate what i said

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Number of Songs Total Cost
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Step-by-step explanation:

8 0
3 years ago
Jamie has 100 milliliters of a 50% isopropyl alcohol solution. How many milliliters of a 96% isopropyl alcohol solution does she
serious [3.7K]

Answer:

318.2 ml

Step-by-step explanation:

Jamie has 100 ml of a 50% isopropyl alcohol solution.

She wants to prepare a solution of 85% by adding 96% solution of isopropyl alcohol solution.

Let the amount of 96% solution required = x ml

Final amount of 85% isopropyl alcohol solution = (100 + x) ml

Therefore, equation for this situation will be,

100 × (50%) + (x) × (96%) = (100 + x) × (85%)

100(0.5) + 0.96x = 0.85(100 + x)

50 + 0.96x = 85 + 0.85x

0.96x - 0.85x = 85 - 50

0.11x = 35

x = \frac{35}{0.11}

x = 318.18 ml

  ≈ 318.2 ml

Therefore, she will require 318.2 ml of 96% isopropyl alcohol solution.

4 0
3 years ago
Multiples of 4 are even numbers
Vesna [10]

Answer:

The statement is true

Step-by-step explanation:

Multiples of 4 (until 40): 4 , 8 , 12 , 16 , 20 , 24 , 28 , 32 , 36 , 40

All of these are even numbers

4 0
3 years ago
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At 2 PM, a thermometer reading 80°F is taken outside where the air temperature is 20°F. At 2:03 PM, the temperature reading yiel
Alexeev081 [22]
Use Law of Cooling:
T(t) = (T_0 - T_A)e^{-kt} +T_A
T0 = initial temperature, TA = ambient or final temperature

First solve for k using given info, T(3) = 42
42 = (80-20)e^{-3k} +20 \\  \\ e^{-3k} = \frac{22}{60}=\frac{11}{30} \\  \\ k = -\frac{1}{3} ln (\frac{11}{30})

Substituting k back into cooling equation gives:
T(t) = 60(\frac{11}{30})^{t/3} + 20

At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\  \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80
Solve for x:
x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80
Sub back into original cooling equation, x = T(t)
-9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20
Solve for t:
60 (\frac{11}{30})^{t/3} +9(\frac{11}{30})^{\frac{-10}{3}}(\frac{11}{30})^{t/3}  = 60 \\  \\  (\frac{11}{30})^{t/3} = \frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}} \\  \\  \\ t = 3(\frac{ln(\frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}})}{ln (\frac{11}{30})}) \\  \\ \\  t = 4.959

This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM
8 0
3 years ago
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