Add 531+642+753+864+975

Need help!!! no idea what to do?!<br><br>plz help

vagabundo [1.1K]

Which question was it that you were wanting me to answer?

8 0

3 years ago

Can someone help me with this question?

kap26 [50]

$\frac{4x + 5 - 5x}{ {6x}^{2} - x - 12 } \\ = \frac{ - x + 5}{ {6x}^{2} - x - 12}$

The answer is the first one.

Hope this helps. - M

4 0

3 years ago

Samantha reports that 60% of the first year students at her university think they should be able to bring a car to campus. She a

Nostrana [21]

Answer:

D) the standard error for her sample is 15

Step-by-step explanation:

Samantha is 95% certain that between 50% and 70% of the first year students agree that they should be able to bring a car to campus.

Here

95% is the confidence level.
The range between 50% and 70% indicates 95% confidence interval.
60% is the average proportion of the first year students at her university think they should be able to bring a car to campus.
10% (0.1) is the margin of error from the mean

Since standard error can be obtained by dividing margin of error by the z-statistic of the confidence level which is 1.96. The standard error for her sample $\frac{0.1}{1.96}$ does not equal to 15.

4 0

3 years ago

Tacoma's population in 2000 was about 200 thousand, and had been growing by about 9% each year. a. Write a recursive formula for

KIM [24]

Answer:

a) The recurrence formula is $P_n = \frac{109}{100}P_{n-1}$ .

b) The general formula for the population of Tacoma is

$P_n = \left(\frac{109}{100}\right)^nP_{0}$ .

c) In 2016 the approximate population of Tacoma will be 794062 people.

d) The population of Tacoma should exceed the 400000 people by the year 2009.

Step-by-step explanation:

a) We have the population in the year 2000, which is 200 000 people. Let us write $P_0 = 200 000$ . For the population in 2001 we will use $P_1$ , for the population in 2002 we will use $P_2$ , and so on.

In the following year, 2001, the population grow 9% with respect to the previous year. This means that $P_0$ is equal to $P_1$ plus 9% of the population of 2000. Notice that this can be written as

$P_1 = P_0 + (9/100)*P_0 = \left(1-\frac{9}{100}\right)P_0 = \frac{109}{100}P_0$ .

In 2002, we will have the population of 2001, $P_1$ , plus the 9% of $P_1$ . This is

$P_2 = P_1 + (9/100)*P_1 = \left(1-\frac{9}{100}\right)P_1 = \frac{109}{100}P_1$ .

So, it is not difficult to notice that the general recurrence is

$P_n = \frac{109}{100}P_{n-1}$ .

b) In the previous formula we only need to substitute the expression for $P_{n-1}$ :

$P_{n-1} = \frac{109}{100}P_{n-2}$ .

Then,

$P_n = \left(\frac{109}{100}\right)^2P_{n-2}$ .

Repeating the procedure for $P_{n-3}$ we get

$P_n = \left(\frac{109}{100}\right)^3P_{n-3}$ .

But we can do the same operation n times, so

$P_n = \left(\frac{109}{100}\right)^nP_{0}$ .

c) Recall the notation we have used:

$P_{0}$ for 2000, $P_{1}$ for 2001, $P_{2}$ for 2002, and so on. Then, 2016 is $P_{16}$ . So, in order to obtain the approximate population of Tacoma in 2016 is