Hey can yall help me in dis rq

A swimming pool is being drained so it can be cleaned. The amount of water in the pool is changing according to the función f(t)

TiliK225 [7]

Answer:

0 ≤ t ≤ 5.

Step-by-step explanation:

In the function $f(t)$ , $t$ is the independent variable. The domain of $f$ is the set of all values of $t$ that this function can accept.

In this case, $f(t)$ is defined in a real-life context. Hence, consider the real-life constraints on the two variables. Both time and volume should be non-negative. In other words,

$t \ge 0$ .
$f(t) \ge 0$ .

The first condition is an inequality about $t$ , which is indeed the independent variable.

However, the second condition is about $f$ , the dependent variable of this function. It has to be rewritten as a condition about $t$ .

$\begin{aligned} f(t) &\ge 0 &&\text{Assumption} \cr 80000 - 16000\, t& \ge 0 && \text{Definition of} ~ f \cr 80000 & \ge 16000\, t && \begin{aligned}&\text{Add $16000\, t$} \\[-0.5em] & \text{to both sides of the inequality}\end{aligned} \cr 5 &\ge t &&\begin{aligned}&\text{Divide both sides of} \\[-0.5em] & \text{the inequality by $16000$}\end{aligned} \cr t &\le 5 && \text{Flip the inequality}\end{aligned}$ .

Hence, t ≤ 5.

Combine the two inequalities to obtain the domain:

0 ≤ t ≤ 5.

6 0

2 years ago

Because of staffing decisions, managers of the Gibson-Marion Hotel are interested in the variability in the number of rooms occu

olchik [2.2K]

Answer:

a) $s^2 =30^2 =900$

b) $\frac{(19)(30)^2}{30.144} \leq \sigma^2 \leq \frac{(19)(30)^2}{10.117}$

$567.28 \leq \sigma^2 \leq 1690.224$

c) $23.818 \leq \sigma \leq 41.112$

Step-by-step explanation:

Assuming the following question: Because of staffing decisions, managers of the Gibson-Marimont Hotel are interested in the variability in the number of rooms occupied per day during a particular season of the year. A sample of 20 days of operation shows a sample mean of 290 rooms occupied per day and a sample standard deviation of 30 rooms

Part a

For this case the best point of estimate for the population variance would be:

$s^2 =30^2 =900$

Part b

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

The degrees of freedom are given by:

$df=n-1=20-1=19$

Since the Confidence is 0.90 or 90%, the significance $\alpha=0.1$ and $\alpha/2 =0.05$ , the critical values for this case are:

$\chi^2_{\alpha/2}=30.144$

$\chi^2_{1- \alpha/2}=10.117$

And replacing into the formula for the interval we got:

$\frac{(19)(30)^2}{30.144} \leq \sigma^2 \leq \frac{(19)(30)^2}{10.117}$

$567.28 \leq \sigma^2 \leq 1690.224$

Part c

Now we just take square root on both sides of the interval and we got:

$23.818 \leq \sigma \leq 41.112$

5 0

3 years ago

What is 472.746 rounded to nearest whole number ?

lorasvet [3.4K]

500.000 is your answer

4 0

2 years ago

4. Mario was asked to find the GCF of 15, 30, and 45. What number did he find? A. 15 C. 45 B. 30 D. 60

SVETLANKA909090 [29]

4. is A 15
5. is C 12
6. is B 4, 8, 12, 16

hope this helps :)

7 0

2 years ago

Read 2 more answers

Which of the scatterplots to the right show a) no association? b) a negative association? c) a linear association? d) a w

Len [333]

Answer:

A scatter plot to the right shows a very strong association.

Step-by-step explanation:

Because a scatter plot to the right shows that both variables are positive and they both increase, so it shows a strong association between those variables.

4 0

3 years ago