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Ivanshal [37]
3 years ago
8

The Summer Olympics were held in St. Louis 92 years before the Summer Olympics were held in Atlanta. In what year were the Olymp

ics held in St. Louis? 1996 Atlanta u.s 2000 Sydney Australia 2004 Athens Greece 2008 Beijing china 2012 london Great Britain.
Mathematics
2 answers:
andrezito [222]3 years ago
7 0
1904 write if not just subtract 92 from 96 and you have your answer hope that helps
Rzqust [24]3 years ago
5 0
1904 is the answer
Just subtract 92 from 1996
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At the carnival, a ferris wheel ride costs 2 tickets and a roller coaster ride costs 4 tickets. Miles has 16 tickets. Write the
Alexxandr [17]

Answer:

2F + 4R = 16

Step-by-step explanation:

F: no. of ferris wheel rides

R: no. of roller coaster rides

2F + 4R = 16

4 0
3 years ago
The temperature of a certain solution is estimated by taking a large number of independent measurements and averaging them. The
Mademuasel [1]

Answer:

(a) The 95% confidence interval for the temperature is (36.80°C, 37.20°C).

(b) The confidence level is 68%.

(c) The necessary assumption is that the population is normally distributed.

(d) The 95% confidence interval for the temperature if 10 measurements were made is (36.93°C, 37.07°C).

Step-by-step explanation:

Let <em>X</em> = temperature of a certain solution.

The estimated mean temperature is, \bar x=37^{o}C.

The estimated standard deviation is, s=0.1^{o}C.

(a)

The general form of a (1 - <em>α</em>)% confidence interval is:

CI=SS\pm CV\times SD

Here,

SS = sample statistic

CV = critical value

SD = standard deviation

It is provided that a large number of independent measurements are taken to estimate the mean and standard deviation.

Since the sample size is large use a <em>z</em>-confidence interval.

The critical value of <em>z</em> for 95% confidence interval is:

z_{0.025}=1.96

Compute the confidence interval as follows:

CI=SS\pm CV\times SD\\=37\pm 1.96\times 0.1\\=37\pm0.196\\=(36.804, 37.196)\\\approx (36.80^{o}C, 37.20^{o}C)

Thus, the 95% confidence interval for the temperature is (36.80°C, 37.20°C).

(b)

The confidence interval is, 37 ± 0.1°C.

Comparing the confidence interval with the general form:

37\pm 0.1=SS\pm CV\times SD

The critical value is,

CV = 1

Compute the value of P (-1 < Z < 1) as follows:

P(-1

The percentage of <em>z</em>-distribution between -1 and 1 is, 68%.

Thus, the confidence level is 68%.

(c)

The confidence interval for population mean can be constructed using either the <em>z</em>-interval or <em>t</em>-interval.

If the sample selected is small and the standard deviation is estimated from the sample, then a <em>t</em>-interval will be used to construct the confidence interval.

But this will be possible only if we assume that the population from which the sample is selected is Normally distributed.

Thus, the necessary assumption is that the population is normally distributed.

(d)

For <em>n</em> = 10 compute a 95% confidence interval for the temperature as follows:

The (1 - <em>α</em>)% <em>t</em>-confidence interval is:

CI=\bar x\pm t_{\alpha/2, (n-1)}\times \frac{s}{\sqrt{n}}

The critical value of <em>t</em> is:

t_{\alpha/2, (n-1)}=t_{0.025, 9}=2.262

*Use a <em>t</em>-table for the critical value.

The 95% confidence interval is:

CI=37\pm 2.262\times \frac{0.1}{\sqrt{10}}\\=37\pm 0.072\\=(36.928, 37.072)\\\approx (36.93^{o}C, 37.07^{o}C)

Thus, the 95% confidence interval for the temperature if 10 measurements were made is (36.93°C, 37.07°C).

3 0
2 years ago
HELP ME PLZ ILL GIVE YOU EXTRA POINTS IF U GET IT RIGHT
tamaranim1 [39]

Step-by-step explanation:

I used my calculator, so they should be correct.

6 0
3 years ago
Read 2 more answers
Use breaking apart to find the product for 6x135
-BARSIC- [3]
135x6 equals 810 thAt is the answer
8 0
3 years ago
You have a new pool and want to know its volume. The pool is 5 feet deep and has a radius of 7 feet. How much water can a pool h
leva [86]

Answer:

769.6902 (or 769.3) cubic feet of water

Step-by-step explanation:

πr^2

r = 7

5 x π(7)^2 = 769.6902 (or 769.3 using 3.14)

5 0
2 years ago
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