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FrozenT [24]
3 years ago
8

Help plz!!,!!!!!!,!!!!!,!,!,!,

Mathematics
1 answer:
borishaifa [10]3 years ago
6 0

Answer:C

Step-by-step explanation:

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Mr. Winters has a $50 daily budget for car rental. He has rented a car for a flat $10 a day, plus $0.12 per mile. What mileage w
daser333 [38]

Answer:

333 miles

Step-by-step explanation:

50 = 10 + 0.12x

40 = 0.12x

333.33 miles = x

equals to spending $49.96

8 0
2 years ago
I don’t know how to solve #47. What steps do I do to find m?
grandymaker [24]

Hope this helps :).  

5 0
3 years ago
Exponential function
laiz [17]
A typical exponential function is y=ab^{x}
when x=3.5, y=16.2
when x=6, y=3936.6
plug these values into the exponential function:
16.2=ab^{3.5}
3936.6=ab^{6}
divide the second equation by the first to eliminate a:
243=b^{(6-3.5)}
log both sides: log243=2.5logb
logb=log243/2.5
use your calculator to find b: b=3.6
plug b=3.6 in the first equation to find a:
16.2=a* (3.6)^{3.5}
a=0.183

please double check my calculation
5 0
3 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
Please what the answer is?
gulaghasi [49]

Answer:

26

Step-by-step explanation:

Took test

3 0
2 years ago
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