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3 years ago

5 over 8- (1 over 4) to the 2nd power equals

Mathematics

1 answer:

vlabodo [156]3 years ago

7 0

Answer:

0.5625. this is what it equals

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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: $F=0.055 L\s$

4 0

3 years ago

For this set of data,<br> 89, 88, 84, 87, 89, 84, 84, 83, 85, 89, 84<br> Find the standard deviation

Ilia_Sergeevich [38]

Answer: 2.2962419891482

Step-by-step explanation: First get the mean of the numbers or the average

the mean in this case is 86

then subtract each number by the mean of 86

the numbers then become : 3, 2, -2, 1, 3, -2, -2, -3, -1, 3, -2

then square all the numbers to get : 9, 4, 4, 1, 9, 4, -4, 9, 1, 9, 4

then find the mean of those numbers which is 5.2727272727273

then find the square root of this number to find the standard deviation

2.2962419891482 is the standard deviation

hope this helps mark me brainliest if it helped

6 0

3 years ago

0.5s +1=7 + 4.5s<br> S =

kvv77 [185]

Answer:

$s = -\frac{3}{2}$

Step-by-step explanation:

Solve for the value of $s$ :

$0.5s + 1 = 7 + 4.5s$

-Combine both $0.5s$ and $4.5s$ by subtracting $0.5s$ by $4.5s$ :

$0.5s + 1 - 4.5s = 7 + 4.5s - 4.5s$

$-4s + 1 = 7$

-Subtract $1$ to both sides:

$-4s + 1 - 1 = 7 - 1$

$-4s = 6$

-Divide both sides by $-4$ :

$\frac{-4s}{-4} = \frac{6}{-4}$

$s = -\frac{3}{2}$

So, the value of $s$ is $-\frac{3}{2}$ .

3 0

3 years ago

The sum of two consecutive numbers is 5. Find the equation representing<br> the same.

serious [3.7K]

Answer:

Step-by-step explanation:

Let the smallest number = x

The next number will be x + 1

x + x + 1 = 5

2x + 1 = 5 This is likely the answer to the question although it could be the equation above.

7 0

2 years ago

Suppose 44% of the children in a school are girls. If a sample of 727 children is selected, what is the probability that the sam

Harman [31]

Answer:

0.9484 = 94.84% probability that the sample proportion of girls will be greater than 41%

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$