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3 years ago

How to find the average of 78, 79, 87, 88, 101, 120, 130

Mathematics

2 answers:

crimeas [40]3 years ago

8 0

Answer:

Add the numbers together and divide by the number of numbers. (The sum of values divided by the number of values).

Step-by-step explanation:

Mkey [24]3 years ago

7 0

Answer:

97.5714286

Step-by-step explanation:

1. You would have to add all the numbers together

ex. 78 + 79 + 87 + 88 + 101 + 120 + 130 = 683

2. Then divid the sum by the number of numbers added.

ex. 683/7 = 97.5714286

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Consider a binary code with 5 bits (0 or 1) in each code word. An example of a code word is 01001. In each code word, a bit is z

AleksandrR [38]

Answer:

the probability that a code word contains exactly one zero is 0.0064 (0.64%)

Step-by-step explanation:

Since each bit is independent from the others , then the random variable X= number of 0 s in the code word follows a binomial distribution, where

p(X)= n!/((n-x)!*x!*p^x*(1-p)^(n-x)

where

n= number of independent bits=5

x= number of 0 s

p= probability that a bit is 0 = 0.8

then for x=1

p(1) = n*p*(1-p)^(n-1) = 5*0.8*0.2^4 = 0.0064 (0.64%)

therefore the probability that a code word contains exactly one zero is 0.0064 (0.64%)

3 0

3 years ago

What is the perimeter of this rectangle?

Natali [406]

d. you add up all the sides. opposite sides are equal. the answer is d. 36

7 0

2 years ago

Suppose that a box contains one fair coin (Heads and Tails are equally likely) and one coin with Heads on each side. Suppose fur

stealth61 [152]

Using conditional probability, it is found that there is a 0.1 = 10% probability that the chosen coin was the fair coin.

Conditional Probability

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.
$P(A \cap B)$ is the probability of both A and B happening.
P(A) is the probability of A happening.

In this problem:

Event A: Three heads.
Event B: Fair coin.

The probability associated with 3 heads are:

$0.5^3 = 0.125$ out of 0.5(fair coin).
1 out of 0.5(biased).

Hence:

$P(A) = 0.125 + 0.5 = 0.625$

The probability of 3 heads and the fair coin is:

$P(A \cap B) = 0.5(0.125) = 0.0625$

Then, the conditional probability is:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.0625}{0.625} = 0.1$

0.1 = 10% probability that the chosen coin was the fair coin.

A similar problem is given at brainly.com/question/14398287

4 0

2 years ago

1. Use synthetic division to determine each quotient. Show your work.

Brut [27]

A= x^2+2x-6
B= x^3-x+5-(28/x+3)

5 0

2 years ago

Which point approximates √30? A) A B) B C) C D) D

Paha777 [63]

Letter D

Hope this helps

3 0

2 years ago

Read 2 more answers