Fill in the blanks

a big cruise ship dropped anchor off the Caribbean island of antigua. the heavy anchor dropped into the water at a rate of 2.5 m

AysviL [449]

Falling at the rate of 2.5 m/s for 45 seconds, the anchor fell

(2.5 m/s) x (45 s) = 112.5 meters .

If it wound up only 40 meters underwater, then it must have fallen

(112.5 m - 40 m) = 72.5 meters

from its storage position on the big cruise ship,
before it ever reached the water.

Since it was falling at the rate of 2.5 m/s, it took

(72.5 m) / (2.5 m/s) = 29 seconds

to get down to the water surface after it started dropping.

6 0

2 years ago

A number is shown in expanded form.

Nana76 [90]

20,809,204 is the answer because 20,000,000+800,000+9,000+200+4

4 0

3 years ago

Read 2 more answers

100-(2 8/5+1 3/8)×(3.05-0.8

Leya [2.2K]

<span>88.80625 is the answer because I used a calculator.</span>

5 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

32) (a)-(c) pre calculus

Harrizon [31]

Well, remember we can't take the square root of a negative

so we see that we have $\sqrt{1-x^2}$
so find those values that take sqrt of a negative and restrict hem from the domain
anny value greater than 1 and less than -1
so domain is from -1 to 1, including those numbers
D=[-1,1]

a. D=[-1,1] or from -1 to 1 is domain

b. for a TI-84, go to y-editor then input $4x \sqrt{1-x^2}$ for y1

c. for a TI-84, click 2nd then window (gets to tbset) scrol down to set Δx to 0.1, then cilick 2nd again then click graph (to select table) and scroll down till you see that value of y that is the biggest, that value is x=0.7

A. domain is from -1 to 1
B. use your brain or google the instructions for your calulator
C. at x=0.7

8 0

3 years ago