10=1<br> bbb nbththhhhhhhhhhhhhhhhhhhhhhh

6sqrt20 + 8sqrt5 - 5sqrt45

marta [7]

Answer:

11.1803398875

Step-by-step explanation:

4 0

3 years ago

I need help with number 1

Ivan

The answer you have is good. You don't need to change anything.

3 0

3 years ago

the half life of c14 is 5730 years. Suppose that wood found at an archeological excavation site contains about 35% as much C14 a

Furkat [3]

Answer:

The wood was cut approximately 8679 years ago.

Step-by-step explanation:

At first we assume that examination occured in 2020. The decay of radioactive isotopes are represented by the following ordinary differential equation:

$\frac{dm}{dt} = -\frac{m}{\tau}$ (Eq. 1)

Where:

$\frac{dm}{dt}$ - First derivative of mass in time, measured in miligrams per year.

$\tau$ - Time constant, measured in years.

$m$ - Mass of the radioactive isotope, measured in miligrams.

Now we obtain the solution of this differential equation:

$\int {\frac{dm}{m} } = -\frac{1}{\tau}\int dt$

$\ln m = -\frac{1}{\tau} + C$

$m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }$ (Eq. 2)

Where:

$m_{o}$ - Initial mass of isotope, measured in miligrams.

$t$ - Time, measured in years.

And time is cleared within the equation:

$t = -\tau \cdot \ln \left[\frac{m(t)}{m_{o}} \right]$

Then, time constant can be found as a function of half-life:

$\tau = \frac{t_{1/2}}{\ln 2}$ (Eq. 3)

If we know that $t_{1/2} = 5730\,yr$ and $\frac{m(t)}{m_{o}} = 0.35$ , then:

$\tau = \frac{5730\,yr}{\ln 2}$

$\tau \approx 8266.643\,yr$

$t = -(8266.643\,yr)\cdot \ln 0.35$

$t \approx 8678.505\,yr$

The wood was cut approximately 8679 years ago.

5 0

3 years ago

Two opposing opinions were shown to a random sample of 1,500 buyers of a particular political news app in the United States. The

kifflom [539]

Answer:

A) 98% Confidence interval for the proportion of all US buyers of this particular app who would have chosen Opinion B

= (0.51, 0.57)

This means that the true proportion of all thay would chose opinion B can take on values between the range of (0.51, 0.57)

B) For the confidence interval obtained to be valid, the conditions stated must be satisfied and for the sampling distribution to be approximately normal, the number of buyers that chose Opinion B and the number of buyers that did not choose Opinion B must both be greater than 10.

Step-by-step explanation:

Confidence Interval for the population proportion is basically an interval of range of values where the true population proportion can be found with a certain level of confidence.

Mathematically,

Confidence Interval = (Sample proportion) ± (Margin of error)

Sample proportion of all US buyers of this particular app who would have chosen Opinion B = 0.54

Margin of Error is the width of the confidence interval about the mean.

It is given mathematically as,

Margin of Error = (Critical value) × (standard Error)

Critical value at 98% confidence interval for sample size of 1500 is obtained from the z-tables.

Critical value = 2.33

Standard error = σₓ = √[p(1-p)/n]

p = sample proportion = 0.54

n = sample size = 1500

σₓ = √(0.54×0.46/1500) = 0.0128685664 = 0.01287

98% Confidence Interval = (Sample proportion) ± [(Critical value) × (standard Error)]

CI = 0.54 ± (2.33 × 0.01287)

CI = 0.54 ± 0.02998)

98% CI = (0.51, 0.57)

98% Confidence interval = (0.51, 0.57)

B) The number of buyers that chose Opinion B and the number of buyers that did not choose Opinion B are both greater than 10. Why must this inference condition be met?

For this confidence interval to be obtained using sample data, a couple of conditions are necessary to be satisfied. They include;

- The sample data must have been obtained using a random sampling technique.

- The sampling distribution must be normal or approximately normal.

- The variables of the sample data must be independent of each other.

On the second point, the condition for a binomial distribution to approximate a normal distribution is that

np ≥ 10

and n(1-p) ≥ 10

The quantity np is the actual sample mean which is the actual number of buyers that chose Opinion B while n(1-p) is the number of buyers that did not chose Opinion B.

For the confidence interval obtained to be valid, the conditions stated must be satisfied and for the sampling distribution to be approximately normal, the number of buyers that chose Opinion B and the number of buyers that did not choose Opinion B must both be greater than 10.

Hope this Helps!!!

8 0

3 years ago

What is the slope of the line that passes through the points<br> (3, -7) and (-5, 17)?

MatroZZZ [7]

-3 is the slope, use y2 - y1 over x2 - x1 equation to solve

4 0

3 years ago

10=1 bbb nbththhhhhhhhhhhhhhhhhhhhhhh