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wariber [46]
3 years ago
8

( PLEASE ANSWER WILL GIVE BRAINIEST AND A THANK YOU ON PROFILE!!) On the previous problem, you found a unit rate of ounces per b

ox. Explain how you find a unit rate when given a rate.
Mathematics
2 answers:
kari74 [83]3 years ago
8 0
<span>First, I would write the rate as a fraction. Then I would divide the numerator by the denominator.</span>
Nostrana [21]3 years ago
4 0
To find<span> the </span>unit rate<span>, divide the numerator and denominator of the </span>given rate<span> by the denominator of the </span>given rate<span>. So in this case, divide the numerator and denominator of 70/5 by 5, to </span>get<span> 14/1, or 14 students per class, which is the </span>unit rate<span>.</span>
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the sum of 49,000 naira is to be shared among A, B, and C in the ratio of 1:2:4, respectively. find how much each wilk receive.​
Natalka [10]

Answer:

#7000, #14000 and #28000 respectively

Step-by-step explanation:

#49000 to be shared in 1:2:4

49000 ÷ (1 + 2 + 4) = 49000/7 = #7000

A will get (7000×1) = #7000

B will get (7000×2) = #14000

C will get (7000×4) = #28000

3 0
3 years ago
You’ve observed the following returns on Crash-n-Burn Computer’s stock over the past five years: 14 percent, –9 percent, 16 perc
MatroZZZ [7]

Answer:

a. 9%

b. 0.01445

c. 12.02%

Step-by-step explanation:

n = 5 years

Return rates = 14%, –9%, 16%, 21%, and 3%

a. Arithmetic average return

A= \frac{14-9+16+21+3}{5} \\A=9.00

b. Historical variance (to 5 decimal places)

V= \frac{\sum (X_i-A)^2}{n-1} \\V= \frac{(0.14-0.09)^2+(-0.09-0.09)^2+(0.16-0.09)^2(0.21-0.09)^2(0.03-0.09)^2}{5-1}\\ V=0.01445

c. Standard deviation.

s = \sqrt{V} \\s= \sqrt{0.01445}\\s=0.1202 = 12.02\%

8 0
3 years ago
1. How much heat is absorbed by a 50g iron skillet when its temperature rises from 10oC to 124oC? Joules
hoa [83]
<span>1. How much heat is absorbed by a 50g iron skillet when its temperature rises from 10oC to 124oC? Joules

Formula: Heat = mass * specific heat * ΔT

Data:
mass = 50 g = 0.050 kg
specific heat of iron = 450 J/ kg °C
ΔT = 124°C - 10°C ¿ 114 °C

=> heat = 0.050kg * 450 J / kg°C * 114°C ≈ 2.6 J

2. If a refrigerator is a heat pump that follows the first law of thermodynamics, how much heat was removed from food inside of the refrigerator if it released 492J of energy to the room? Joules

The firs law of thermodynamics is conservation of energy => energy removed from inside of the refrigerator = energy released to the room

=> Answer = 492 J

3. How much heat is needed to raise the temperature of 45g of water by 63oC? Joules

Formula: heat = mass * specific heat * ΔT

specific heat of water = 4186 J / Kg °C

heat = 0.045 kg * 4186 J/kg °C * 63°C = 11,867.31 J
</span>
7 0
3 years ago
18 divide by 2 * 3/5-2 simplify show all work
liraira [26]

Answer:

9

Step-by-step explanation:

I would guess it looks like this - if not then tell me

\frac{18}{ \frac{2 \times 3}{5 - 2} }

2×3=6

5-2=3

Then there's only

\frac{18}{ \frac{6}{3} }

Which is equal to

\frac{18}{1}  \times  \frac{3}{6}  =  \frac{54}{6}  =  \frac{9}{1}  = 9

6 0
3 years ago
Read 2 more answers
Theorem 8.9 : The line segment joining the mid-points of two sides of a triangle is parallel to the third side.
Alla [95]

Answer:

The essence of Mathematics lies in its freedom

7 0
3 years ago
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