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3 years ago

Complete the equation:1(2/3) mi =how many ft.

1 answer:

4 0

There are 5280 feet in one mile.

If you have one and two-thirds of a mile, you can divide 5280 by 3, multiply it by 2, and then add it to 5280.

This will give you an answer of 8800 feet in 1 2/3 miles

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Unit 3 Exam

lbvjy [14]

Answer:

Scalene is the right option.

Step-by-step explanation:

A scalene triangle has all its sides of different lengths.

6 0

3 years ago

A study investigated the effectiveness of meditation training in reducing trait anxiety. The study evaluated subjects trait anxi

Setler79 [48]

Answer:

The 99% confidence interval for the population mean reduction in anxiety was (1.2, 8.6).

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 27 - 1 = 26

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 26 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.99}{2} = 0.995$ . So we have T = 2.7787.

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 2.7787\frac{6.9}{\sqrt{27}} = 3.7$

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 4.9 - 3.7 = 1.2.

The upper end of the interval is the sample mean added to M. So it is 4.9 + 3.7 = 8.6.

The 99% confidence interval for the population mean reduction in anxiety was (1.2, 8.6).

6 0

2 years ago

The smaller triangle was dilated to form the larger triangle what is the value of x

Sladkaya [172]

Answer:

i think c which is 14

3 0

3 years ago

The Department of Agriculture is monitoring the spread of mice by placing 100 mice at the start of the project. The population,

uranmaximum [27]

Answer:

Step-by-step explanation:

Assuming that the differential equation is

$\frac{dP}{dt} = 0.04P\left(1-\frac{P}{500}\right)$ .

We need to solve it and obtain an expression for $P(t)$ in order to complete the exercise.

First of all, this is an example of the logistic equation, which has the general form

$\frac{dP}{dt} = kP\left(1-\frac{P}{K}\right)$ .

In order to make the calculation easier we are going to solve the general equation, and later substitute the values of the constants, notice that $k=0.04$ and $K=500$ and the initial condition $P(0)=100$ .

Notice that this equation is separable, then

$\frac{dP}{P(1-P/K)} = kdt$ .

Now, intagrating in both sides of the equation

$\int\frac{dP}{P(1-P/K)} = \int kdt = kt +C$ .

In order to calculate the integral in the left hand side we make a partial fraction decomposition:

$\frac{1}{P(1-P/K)} = \frac{1}{P} - \frac{1}{K-P}$ .

So,

$\int\frac{dP}{P(1-P/K)} = \ln|P| - \ln|K-P| = \ln\left| \frac{P}{K-P} \right| = -\ln\left| \frac{K-P}{P} \right|$ .

We have obtained that:

$-\ln\left| \frac{K-P}{P}\right| = kt +C$

which is equivalent to

$\ln\left| \frac{K-P}{P}\right|= -kt -C$

Taking exponentials in both hands:

$\left| \frac{K-P}{P}\right| = e^{-kt -C}$

Hence,

$\frac{K-P(t)}{P(t)} = Ae^{-kt}$ .

The next step is to substitute the given values in the statement of the problem:

$\frac{500-P(t)}{P(t)} = Ae^{-0.04t}$ .

We calculate the value of $A$ using the initial condition $P(0)=100$ , substituting $t=0$ :

$\frac{500-100}{100} = A}$ and $A=4$ .

So,

$\frac{500-P(t)}{P(t)} = 4e^{-0.04t}$ .

Finally, as we want the value of $t$ such that $P(t)=200$ , we substitute this last value into the above equation. Thus,

$\frac{500-200}{200} = 4e^{-0.04t}$ .

This is equivalent to $\frac{3}{8} = e^{-0.04t}$ . Taking logarithms we get $\ln\frac{3}{8} = -0.04t$ . Then,

$t = \frac{\ln\frac{3}{8}}{-0.04} \approx 24.520731325$ .

So, the population of rats will be 200 after 25 months.

6 0

3 years ago

He counted 206 wheels and 170 pedals. How many bicycles and tricycles does he have.

noname [10]

Bicycle has 2 wheels and 2 pedals
tricycle has 3 wheels and 2 pedals

EQ 1 : 2b + 2t = 170 pedals

EQ 2: 2b +3t = 206 wheels

subtract EQ 1 from EQ 2

2b +3t = 206 - 2b + 2t = 170 = t=36

there were 36 tricycles

36 x 2 = 72 pedals

170 pedals - 72 = 98 pedals left
98/2 = 49 bicycles

36 Tricycles and 49 bicycles

8 0

3 years ago