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Alex_Xolod [135]
3 years ago
5

Draw segment EF so that is bisects RS. Mark their intersection appoint A

Mathematics
1 answer:
Neko [114]3 years ago
4 0

Step-by-step explanation:

EF is a segment so it's a line with the end points as E and F. Line RS bisects(cuts in half) line EF. Then where the two lines meet is where A is.

Hope this helped.

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What 3/4 reduced down too
kirill115 [55]

3/4 can't be reduced anymore.

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Read 2 more answers
Jake sold 28 tickets to the the school fair and Jeanie sold 21 tickets. What is the ratio in the simplest for of the number of t
HACTEHA [7]
A. 3/4, you just have to simplify 21/28
4 0
3 years ago
Glenn bought 3 pounds of tomatoes. He used § of them to make sauce.
Darya [45]

Answer:i dont kno what number u put

Step-by-step explanation:

8 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
A.8.4 t=_______kg
Art [367]
<h2><u>ANS</u></h2>

A) 8400 kg

B) 10 km

C) 0.231 kg

D) 6000000 g

E) 0.007 t

F) 0.00125 kg

G) 3900 kg

H) 2 kg

I) 5000 g

J) 3.6 kg

<h3><u>T</u><u>hank</u> <u>You</u> !!!</h3>

7 0
2 years ago
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