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Vinil7 [7]
2 years ago
5

The total number of shares in Marks brothers is increased by 12.8% how many shares will there be after the increase

Mathematics
1 answer:
Lisa [10]2 years ago
8 0

Answer:

Total number of shares is to increase by 12.8%

Step-by-step explanation:

After increase the number of shares will be 87984. hope this helps you :)

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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

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1 year ago
The school playground is in the shape of a pentagon. There is a drinking fountain at each of the 5 corners of the playground. Th
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The answer is 21 save
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A survey of the students in Lance’s school found that 58% of the respondents want the school year lengthened, while 42% think it
alina1380 [7]
The margin of error of a given statistic is an amount that is allowed for in case of miscalculation or change of circumstances.

It is usually the radius or half of the width of the confidence interval of that statistic.

Given that a<span> survey of the students in Lance’s school found that 58% of the respondents want the school year lengthened, while 42% think it should remain the same. The margin of error of the survey is ±10%.

This means that 58% </span><span>± 10% of the </span>respondents want the school year lengthened, while 42% <span><span>± 10% think it should remain the same.</span>

Thus, from 48% to 68% </span><span><span>of the respondents want the school year lengthened, while from 32% to 52% <span>think it should remain the same.</span> </span>

Therefore, according to the survey data, at least 32% of students want the duration of the school year to remain unchanged, and at least 48% want the school year to be lengthened.</span>
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A random sample of 28 statistics tutorials was selected from the past 5 years and the percentage of students absent from each on
SVEN [57.7K]

Answer:

Step-by-step explanation:

Hello!

X: number of absences per tutorial per student over the past 5 years(percentage)

X≈N(μ;σ²)

You have to construct a 90% to estimate the population mean of the percentage of absences per tutorial of the students over the past 5 years.

The formula for the CI is:

X[bar] ± Z_{1-\alpha /2} * \frac{S}{\sqrt{n} }

⇒ The population standard deviation is unknown and since the distribution is approximate, I'll use the estimation of the standard deviation in place of the population parameter.

Number of Absences 13.9 16.4 12.3 13.2 8.4 4.4 10.3 8.8 4.8 10.9 15.9 9.7 4.5 11.5 5.7 10.8 9.7 8.2 10.3 12.2 10.6 16.2 15.2 1.7 11.7 11.9 10.0 12.4

X[bar]= 10.41

S= 3.71

Z_{1-\alpha /2}= Z_{0.95}= 1.645

[10.41±1.645*(\frac{3.71}{\sqrt{28} } )]

[9.26; 11.56]

Using a confidence level of 90% you'd expect that the interval [9.26; 11.56]% contains the value of the population mean of the percentage of absences per tutorial of the students over the past 5 years.

I hope this helps!

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Answe

Step-by-step explanation:

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