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dezoksy [38]
4 years ago
6

A store has 80 modems in its inventory, 30 coming from source A and the remainder from source B. Of the modems from source A, 20

% are defective. Of the modems from source B, 8% are defective. Calculate the probability that two out of a random sample of five modems from the store’s inventory are defective. (A) 0.010 (M) 0.078 (X) 0.102 (D) 0.105 (E) 0.125
Mathematics
1 answer:
Lelu [443]4 years ago
7 0

Answer:

A

Step-by-step explanation: Just use divide them and then add

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4(c – 3) <br><br> I need help now
Novosadov [1.4K]

Answer:

4C3 = 4

Step-by-step explanation:

3 0
4 years ago
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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
3 years ago
10.  
garri49 [273]
This one is D
If you add 8 to 5 you get 13, and 3 to 6 you get 9.  And do the same for the other points
7 0
4 years ago
For which equations is 8 a solution? Select the four correct answers.
sveta [45]

Answer:

The answers include

x + 2 =10

x - 4 = 4

3x = 24

Start Fraction x Over 8 End Fraction = 1

Step-by-step explanation:

x + 2 = 10; 8 + 2 = 10

x - 4 = 4; 8 - 4 = 4

3x = 24; 3(8) = 24

Start Fraction x Over 8 End Fraction = 1; 8/8 = 1

5 0
3 years ago
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The scale of a map is 1 cm to 5 km. A farm is represented by a rectangle measuring 1.5 cm by 4 cm. What Is the actual area of th
shusha [124]

Answer:

150 km squared

Step-by-step explanation:

5*1.5=7.5

4*5=20

7.5*20=150

4 0
3 years ago
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