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melamori03 [73]
2 years ago
7

I need help, I don't get this.

Mathematics
1 answer:
salantis [7]2 years ago
3 0

Answer:11 question

Y=5

X=3

Perimeter =16cm

Step-by-step explanation:

2x-4= 2*3=6-4=2

Y-3 = 5-3=2

3x-5=3*3=9-5=4

Y-1=5-1=4

2*(3+5)

=2*8

=16

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A = 3 and c = 12 c/a =
Bess [88]

Answer:

4, because 12/3 is 4 cause 3×4=12

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3 years ago
A mountain climber on MT. Everest is at 29,035 ft above sea level . A stone at the bottom of th dead sea is at 1,349 below sea l
Lana71 [14]
29035 is to the right of 0.
-1349 is to the left of 0.

so, the distance between them:

29035 - (-1349) = 29035+1349 = 30384
7 0
3 years ago
Diane keeps her ribbons in a drawer. She has 3 white ribbons, 6 pink ribbons, 2 yellow ribbons, and 7 blue ribbons. Without look
Akimi4 [234]

Answer:

1/3

Step-by-step explanation:

She has 18 ribbons in total and she has 6 pink ribbons so she has a 6/18 chance or 1/3 chance of pulling a pink ribbon

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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

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