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alina1380 [7]
3 years ago
8

Which inequality is graphed? 1. 2<x2. 2>x3. 2<_x4. 2>_x​

Mathematics
1 answer:
marissa [1.9K]3 years ago
4 0

Answer: 3

Step-by-step explanation:

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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
.After making his first deposit, Paul has $758 in his checking account. The next month,
love history [14]
78 is the answer
If you substract 914-836 that equals 78
And if you substract 836-758=78 too
5 0
2 years ago
Read 2 more answers
Simplify to get the right answer
Musya8 [376]

The correct answer would be: 18+18i

5 0
3 years ago
The Center for Medicare and Medical Services reported that there were 295,000 appeals for hospitalization and other Part A Medic
Ymorist [56]

Answer:

(a) 0.00605

(b) 0.0403

(c) 0.9536

(d) 0.98809

Step-by-step explanation:

We are given that 40% of first-round appeals were successful (The Wall Street Journal, October 22, 2012) and suppose ten first-round appeals have just been received by a Medicare appeals office.

This situation can be represented through Binomial distribution as;

P(X=r)= \binom{n}{r}p^{r}(1-p)^{n-r} ; x = 0,1,2,3,....

where,  n = number of trials (samples) taken = 10

            r = number of success

            p = probability of success which in our question is % of first-round

                   appeals that were successful, i.e.; 40%

So, here X ~ Binom(n=10,p=0.40)

(a) Probability that none of the appeals will be successful = P(X = 0)

     P(X = 0) = \binom{10}{0}0.40^{0}(1-0.40)^{10-0}

                   = 1*0.6^{10} = 0.00605

(b) Probability that exactly one of the appeals will be successful = P(X = 1)

     P(X = 1) = \binom{10}{1}0.40^{1}(1-0.40)^{10-1}

                  = 10*0.4^{1} *0.6^{10-1} = 0.0403

(c) Probability that at least two of the appeals will be successful = P(X>=2)

    P(X >= 2) = 1 - P(X = 0) - P(X = 1)

                     = 1 - \binom{10}{0}0.40^{0}(1-0.40)^{10-0} - \binom{10}{1}0.40^{1}(1-0.40)^{10-1}

                     = 1 - 0.00605 - 0.0403 = 0.9536

(d) Probability that more than half of the appeals will be successful =             P(X > 0.5)

  For this probability we will convert our distribution into normal such that;

   X ~ N(\mu = n*p=4,\sigma^{2}= n*p*q = 2.4)

  and standard normal z has distribution as;

      Z = \frac{X-\mu}{\sigma} ~ N(0,1)

  P(X > 0.5) = P( \frac{X-\mu}{\sigma} > \frac{0.5-4}{\sqrt{2.4} } ) = P(Z > -2.26) = P(Z < 2.26) = 0.98809

3 0
3 years ago
Suppose both factors in a multiplication problem are multiples of 10. Why might the number of zeroes in the product be different
Lelu [443]

Answer:

Multiplying factors whose products are multiples of 10 add to the number of zeros when factors are multiplied as multiples of 10s.

Step-by-step explanation:

Let the first five multiples of ten be

10*1= 10

10*2= 20

10*3=30

10*4=40

10*5= 50

Suppose we chose 20 and 50.

Now multiplying 20 with 50 we get

20*50= 1000

IF we count the total number of zeros in the factors ( 20 and 50) they are 2.

But the number of zeros in the product (1000) are 3.

This is because when we multiply 2 with 5 we get 10 which adds to the existing number of zeros ( i.e 2) and we get a total of 3 zeros.

And multiplying 10 with 50 we get

10*50= 500

IF we count the total number of zeros in the factors ( 10 and 50) they are 2.

But the number of zeros in the product (500) are also 2.

This is because when we multiply 1 with 5 we get 5 which does not add to the existing number of zeros ( i.e 2) and the total number of zeros remain the same.

Similarly multiplying 20 with 30 we get

20*30= 600

IF we count the total number of zeros in the factors ( 20 and 30) they are 2.

But the number of zeros in the product (600) are also 2.

This is because when we multiply 2 with 3 we get 6 which does not have a zero and the total  number of zeros remain the same as in the factors.

So we see that multiplying factors whose products are multiples of 10 add to the number of zeros when factors are multiplied as multiples of 10s.

5 0
3 years ago
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