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LiRa
1 month ago
10

In a test of a printed circuit board using a random test pattern, an array of 16 bits is equally likely to be 0 or 1. Assume the

bits are independent.
(a) What is the probability that all bits are 1s? Round your answer to six decimal places (e.g. 98.765432).
(b) What is the probability that all bits are 0s? Round your answer to six decimal places (e.g. 98.765432).
(c) What is the probability that exactly 8 bits are 1s and 8 bits are 0s? Round your answer to three decimal places (e.g. 98.765).
Mathematics
1 answer:
AnnZ [12.3K]1 month ago
8 0

Answer:

A), B), and C) are clarified below.

Step-by-step explanation:

The inquiry involves using binary digits, employing probabilities that are equal for both conditions, by applying a random test pattern, where the formula is derived from p = q.

Simplifying gives us

P[k] = nCk / 2^n

A. Probability of all bits being 1s

16c16/2^16 = 1/65536

B. Probability of all bits being 0s

16c0/2^16 = 1/65536

C. The probability of having exactly 8 bits as 1s and the other 8 as 0s

16c8/2^16 = 12870/65536 => 0.1963 ≈ 19.63%

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A hypothesis regarding the weight of newborn infants at a community hospital is that the mean is 6.6 pounds. A sample of seven i
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If we compare the p value to a predetermined significance level \alpha=0.05 we conclude that we cannot reject the null hypothesis, indicating insufficient evidence to assert that the true average is different from 6.6 pounds.

The best conclusion is:

a. Failing to reject the null hypothesis.

6 0
1 month ago
(b) Ali uses 11% of his $2345 to buy a television.
Zina [12379]
$257.95
8 0
18 days ago
2m - 6 = 8m then 3m =
Leona [12618]

Respuesta:

3m = -3

Desglose paso a paso:

Nos dan la ecuación

2m - 6 = 8m,

por lo que

2m - 8m = 6,

-6m = 6,

m = 6/(-6),

m = -1.

Así, 3m = -3.

5 0
2 months ago
From an industrial area 70 companies were selected at random and 45 of them were panning for expansion next year. Find 95% confi
Zina [12379]

Answer:

Confidence limit = [52.8%, 75.2%]

Step-by-step explanation:

P=\frac{45}{70}= 0.64

(1-P)=1-0.64=0.36

n= 70

P ± z \sqrt{\frac{P(1-P)}{n} }

where the value z will be sourced from the z-table regarding a 95% confidence interval

1-0.95= 0.05/2= 0.025

0.95+0.025= 0.0975

From the z-table, the value of z linked to 0.0975 is 1.96

0.64 ± 1.96 \sqrt{\frac{0.64*0.36}{70} }

0.64 ± 1.96 (0.057)

0.64 ± 0.112

64%% ± 11.2%

thus, the confidence interval is

64+11.2=75.2%

64-11.2=52.8%

[52.8, 75.2]

8 0
20 days ago
Fourteen runners in a marathon had these race times, in hours,
AnnZ [12381]

Answer:

Attached is the histogram illustrating the marathon runners’ times.

Step-by-step explanation:

The provided data is as follows;

2.21

2.25

2.76

3.1

3.3

3.5

3.6

3.77

3.8

4.23

4.25

4.25

4.6

4.9

From this data, we can determine;

The count of runners finishing between 0 and 1 hour = 0

The count of runners finishing between 1 and 2 hours = 0

The count of runners finishing between 2 and 3 hours = 3

The count of runners finishing between 3 and 4 hours = 6

The count of runners finishing between 4 and 5 hours = 5

Based on these frequencies across the various time ranges, the histogram for the provided data has been constructed and is attached.

7 0
1 month ago
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