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statuscvo [17]
2 years ago
8

Lucy’s chocolate bar is 54% cocoa. If the weight of the chocolate bar is 54 grams, how many grams of cocoa does it contain? Roun

d the answer to the nearest tenth
Mathematics
1 answer:
Masteriza [31]2 years ago
4 0

Answer:

The chocolate bar contains 29.2 grams of cocoa.

Step-by-step explanation:

Given that Lucy’s chocolate bar is 54% cocoa and the weight of the chocolate bar is 54 grams.

Hence, the amount of cocoa in the chocolate bar is given by

54% of 54 grams

54% percentage means 54/100 = 0.54

Hence, the amount of cocoa in the chocolate bar is

0.54\times54\\=29.16

When rounded to the nearest tenth, 29.16 should be equal to 29.2

Hence, the chocolate bar contains 29.2 grams of cocoa.

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Furkat [3]
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8 0
3 years ago
Sam bought an electric iron as a price of 25 less than the original price if the original price of it is $2500 at what price did
wlad13 [49]

Answer:

$2475  or $1875

Step-by-step explanation:

Assuming the question meant to say $25 less than the original price, it would be $2475 because $2500-$25= $2475

However, if the question meant to say that Sam bought the iron for 25% less than the original price, it would be $1875 because 2500*0.25= $625

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4 0
3 years ago
What additional information is needed to prove the triangles are congruent by the SAS Prostulate
VladimirAG [237]
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8 0
2 years ago
The dye dilution method is used to measure cardiac output with 3 mg of dye. The dye concentrations, in mg/L, are modeled by c(t)
Lemur [1.5K]

Answer:

Cardiac output:F=0.055 L\s

Step-by-step explanation:

Given : The dye dilution method is used to measure cardiac output with 3 mg of dye.

To Find : Find the cardiac output.

Solution:

Formula of cardiac output:F=\frac{A}{\int\limits^T_0 {c(t)} \, dt} ---1

A = 3 mg

\int\limits^T_0 {c(t)} \, dt =\int\limits^{10}_0 {20te^{-0.06t}} \, dt

Do, integration by parts

[\int{20te^{-0.6t}} \, dt]^{10}_0=[20t\int{e^{-0.6t} \,dt}-\int[\frac{d[20t]}{dt}\int {e^{-0.6t} \, dt]dt]^{10}_0

[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-20te^{-0.6t}}{0.6}+\frac{20}{0.6}\int {e^{-0.6t} \,dt]^{10}_0

[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-20te^{-0.6t}}{0.6}+\frac{20e^{-0.6t}}{(0.6)^2}]^{10}_{0}

[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-200e^{-6}}{0.6}+\frac{20e^{-6}}{(0.6)^2}]+\frac{20}{(0.60^2}

[\int{20te^{-0.6t}} \, dt]^{10}_0=\frac{20(1-e^{-6}}{(0.6)^2}-\frac{200e^{-6}}{0.6}

[\int{20te^{-0.6t}} \, dt]^{10}_0\sim {54.49}

Substitute the value in 1

Cardiac output:F=\frac{3}{54.49}

Cardiac output:F=0.055 L\s

Hence Cardiac output:F=0.055 L\s

4 0
2 years ago
Roger served 5/8 pound of cracker wich was 2/3 of the entire box what was the weight of the crackers originally in the box
horrorfan [7]
Answer:
original weight of the box = 15 / 16 pounds

Explanation:
Assume that the original amount in the box is x.
We are given that:
5/8 pounds represent 2/3 of the total amount (x).
This can be translated into the following equation:
(2/3) x = 5 / 8

Now, we will solve for x as follows:
(2/3) x = 5 / 8
Multiply both sides by 24 to get rid of the denominators as follows:
(2/3) x * 24 = (5 / 8) * 24
16 x = 15
Divide both sides by 16 to isolate the x as follows:
x = 15 / 16 pounds

Hope this helps :)

8 0
3 years ago
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