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katen-ka-za [31]
3 years ago
8

Decision Making: Matthew received a fax giving him the dimensions of the rectangular living room in his new apartment. The dimen

sions given were 3.76m x 5.49m. He used his calculator to work out the area and got 2.06 m2. Do you think this answer is correct?
Mathematics
1 answer:
horsena [70]3 years ago
5 0

Answer:

No

Step-by-step explanation:

His answer is incorrect.  If we round the dimensions, 3.76 m is almost 4 m, and 5.49 m is almost 5.5 m.  4 × 5.5 = 22.  So his answer should be close to 22 m².

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True or false<br> The solution to the inequality 5x+2&lt;175x+2&lt;17 is x&lt;3x&lt;3.
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I believe it is true. So sorry if it's not.

7 0
3 years ago
the ratio of men and women in a certain factory is 3 to 4. there are 198 men. How many workers are there?
LUCKY_DIMON [66]

3 men

4 women

7 total


3 men/ 7 total = 198 men/ x total

using cross products

3*x = 7 * 198

divide each side by 3

x = 7*198/3

x = 462

There are 462 workers


If you mean 3 men and 1 women for a total of 4 workers when you state a  ratio of men and women in a certain factory is 3 to 4.

3/4 =198/x

Using cross products

3x = 4* 198

Divide each side by 3

3x/3 = 4*198/3

x =264

264 workers


It all depends on how you define ratio of men and women in a certain factory is 3 to 4.   This is incorrect phrasing and I took it to be men to women.   You cannot have a ratio of men and women.




5 0
3 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
3 years ago
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