What are the factors of 3/4 xy

The number of trees in a rainforest decreases each month by 0.5%. The forest currently has 2.5 billion trees. Which expression r

Blababa [14]

Answer:

$y=2.5*10^9(0.005^{12 0}) \\\\$

Step-by-step explanation:

The above question is in the form of an exponential decay. The equation for an exponential decay is given by:

$y=ab^x$

where y and x are variables, b < 1, a is the initial value of y (that is the value of y when x = 0).

Let y represent the number of trees left and x represent the number of months. Given that there is currently 2.5 billion trees, therefore a = 2.5 * 10⁹, b = 0.5% = 0.005. The equations becomes:

$y=2.5*10^9(0.005^x)\\\\After\ ten\ years(x=10*12\ months=120\ months):\\\\y=2.5*10^9(0.005^{12 0}) \\\\$

8 0

3 years ago

Find two consecutive integers whose sum is 373.

nadya68 [22]

Answer:

186 and 187

Step-by-step explanation:

Let n be the first integer.

Then n+1 is the next integer.

We are given their sum is 373.

n+(n+1)=373

Combine like terms:

2n+1=373

Subtract 1 on both sides:

2n=372

Divide both sides by 2:

n=186

So n+1=187.

187

+186

--------

373

4 0

3 years ago

The International Air Transport Association surveys business travelers to develop quality ratings for transatlantic gateway airp

motikmotik

Answer:

The 95% confidence interval for the population mean rating is (5.73, 6.95).

Step-by-step explanation:

We start by calculating the mean and standard deviation of the sample:

$M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{50}(6+4+6+. . .+6)\\\\\\M=\dfrac{317}{50}\\\\\\M=6.34\\\\\\s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{49}((6-6.34)^2+(4-6.34)^2+(6-6.34)^2+. . . +(6-6.34)^2)}\\\\\\s=\sqrt{\dfrac{229.22}{49}}\\\\\\s=\sqrt{4.68}=2.16\\\\\\$

We have to calculate a 95% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=6.34.

The sample size is N=50.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

$s_M=\dfrac{s}{\sqrt{N}}=\dfrac{2.16}{\sqrt{50}}=\dfrac{2.16}{7.071}=0.305$