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barxatty [35]
4 years ago
10

1. The following formula relates the quantities temperature in Celsius (C) and temperature in kelvins (K).

Mathematics
2 answers:
aev [14]4 years ago
5 0
A) T(K)<span> = </span>T(°C)<span> + 273.15
or
</span>T (°C) <span>= </span>T(K)<span> - 273.15
</span>
b) T(K) = 23 + 273.15
    T(K) = 296.15 Kelvin

c) T (°C) = T(K) - 273.15
    T (°C) = 300 - 273.15
    T (°C) = 26.85 °C
So,
300 kelvins > 25 degrees Celsius
Ad libitum [116K]4 years ago
4 0
(b) If the temperature is 23 in degrees Celsius, what is the temperature in kelvins? Show your work.

C = K − 273.15
23 = K - 273.15
296.15 Kelvin

(c) Which is greater, 300 kelvins or 25 degrees Celsius? Show your work.

To compare, we need to convert both to the same units. We do as follows:

300 Kelvin
25+273.15 = 298.15 K

Therefore, 300 Kelvin is larger.
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A circle has a diameter of 6 meters. Which statement about the circumference and area is true?
worty [1.4K]

Answer:

The numerical value of the circumference is greater than the numerical value of the area.

Step-by-step explanation:

This has already been answered

8 0
3 years ago
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5g+3(-7+5g)=1-g<br> g equals what?
Wittaler [7]
5g + 3(-7 + 5g) = 1 - g
5g - 21 + 15g = 1 - g
20g - 21 = 1 - g
20g + g = 1 + 21
21g = 22 
g = 22/21 <===
8 0
3 years ago
Find the equivalent expression 9(2a+1)
Sladkaya [172]
18a+9 is the equivalent expression.
7 0
3 years ago
Find the variance of the following data. Round your answer to one decimal place. x 1 2 3 4 5 P(X=x) 0.3 0.2 0.2 0.1 0.2
Reptile [31]

The variance of a distribution is the square of the standard deviation

The variance of the data is 2.2

<h3>How to calculate the variance</h3>

Start by calculating the expected value using:

E(x) = \sum x* P(x)

So, we have:

E(x) = 1 * 0.3 + 2* 0.2 +3 * 0.2 + 4 * 0.1 + 5 * 0.2

This gives

E(x) = 2.7

Next, calculate E(x^2) using:

E(x^2) = \sum x^2* P(x)

So, we have:

E(x^2) = 1^2 * 0.3 + 2^2* 0.2 +3^2 * 0.2 + 4^2 * 0.1 + 5^2 * 0.2

E(x^2) = 9.5

The variance is then calculated as:

Var(x) = E(x^2) - (E(x))^2

So, we have:

Var(x) = 9.5 - 2.7^2

Var(x) = 2.21

Approximate

Var(x) = 2.2

Hence, the variance of the data is 2.2

Read more about variance at:

brainly.com/question/15858152

4 0
2 years ago
Please can someone help me answer these questions? Would be really grateful if it was done by the 12th of January. Thank you :)
FrozenT [24]
1. 4g= 37-5
 ⇒ 4g= 32
⇒ g= 32/4
⇒ g= 8

2. c= 5*6
⇒ c= 30

3. f+4= 3*5
⇒f+4= 15
⇒ f= 15-4
⇒ f= 11.

4. b+6= 30/5
⇒ b+6= 6
⇒ b= 6-6
⇒ b=0

5. a-5= 60/5
⇒ a-5= 12
⇒ a= 12+5
⇒ a= 17

6. e+2= 45/5
⇒ e+2= 9
⇒ e= 9-2
⇒ e= 7

7. d-5= 12/3
⇒ d-5= 4
⇒ d= 4+5
⇒ d= 9

8. m-4= 16/4
⇒ m-4= 4
⇒ m= 4+4
⇒ m= 8

9. q+6= 30/5
⇒ q+6= 6
⇒ q= 6-6
⇒ q= 0

10. h/3= 3+5
⇒ h/3= 8
⇒ h= 8*3
⇒ h= 24

11. 3c= 3-5
⇒ 3c= -2
⇒ c= -2/3

12. x/3= 6-7
⇒ x/3= -1
⇒ x= -1*3
⇒ x= -3

13. 6p= 2+1
⇒ 6p= 3
⇒ p= 3/6
⇒ p= 1/2

14. 2d-7= 27/3
⇒ 2d-7= 9
⇒ 2d= 9+7
⇒ 2d= 16
⇒ d= 16/2
⇒ d=8

15. b-3= 5/2
⇒ b-3= 2.5
⇒ b= 2.5+3
⇒ b= 5.5

16. y-1= 1/3
⇒ y= 1/3+1
⇒ y= 1 1/3

17. 5v= 9-3
⇒ 5v= 6
⇒ v= 6/5
⇒ v= 1.2

18. n-3 = 3*6
⇒ n-3= 18
⇒ n= 18+3
⇒ n= 21

19. t+10= 1*4
⇒ t+10= 4
⇒ t= 4-10
⇒ t= -6

20. 3c+4= 3*5
⇒ 3c+4= 15
⇒ 3c= 15-4
⇒ 3c= 11
⇒ c= 11/3
⇒ c= 3 2/3

Hope this helps~
6 0
4 years ago
Read 2 more answers
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