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iogann1982 [59]

3 years ago

5

-9+8k=7+4k(if there is no solution,type in ''no solution'') k= Answer

2 answers:

Shtirlitz [24]3 years ago

4 0

Answer:

k = 4

Step-by-step explanation:

-9 +8k = 7+4k

Subtract 4k from each side

-9 + 4k = 7

Add 9 to each side

4k = 16

Divide by 4

k = 4

Plug in the value you got for k:

-9 + 8(4) ? 7 + 4(4)

-9 + 32 ? 7 + 16

23 ? 23

23 = 23

pishuonlain [190]3 years ago

4 0

Answer:

k = 4

Step-by-step explanation:

- 9 + 8k = 7 + 4k

- 9 - 7 = 4k - 8k

- 16 = - 4k

- k = - 16/4

- k = - 4

k = 4

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Sveta_85 [38]

Using the binomial distribution, the probabilities are given as follows:

0.3675 = 36.75% probability that more than 4 weigh more than 20 pounds.

0.1673 = 16.73% probability that fewer than 3 weigh more than 20 pounds.

Since P(X > 7) < 0.05, it would be unusual if more than 7 of them weigh more than 20 pounds.

<h3>What is the binomial distribution formula?</h3>

The formula is:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.

n is the number of trials.

p is the probability of a success on a single trial.

The values of the parameters for this problem are:

n = 10, p = 0.4.

The probability that more than 4 weigh more than 20 pounds is:

$P(X > 4) = 1 - P(X \leq 4)$

In which:

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)$

Then:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.4)^{0}.(0.6)^{10} = 0.0061$

$P(X = 1) = C_{10,1}.(0.4)^{1}.(0.6)^{9} = 0.0403$

$P(X = 2) = C_{10,2}.(0.4)^{2}.(0.6)^{8} = 0.1209$

$P(X = 3) = C_{10,3}.(0.4)^{3}.(0.6)^{7} = 0.2150$

$P(X = 4) = C_{10,4}.(0.4)^{4}.(0.6)^{6} = 0.2502$

Hence:

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0061 + 0.0403 + 0.1209 + 0.2150 + 0.2502 = 0.6325$

$P(X > 4) = 1 - P(X \leq 4) = 1 - 0.6325 = 0.3675$

0.3675 = 36.75% probability that more than 4 weigh more than 20 pounds.

The probability that fewer than 3 weigh more than 20 pounds is:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0061 + 0.0403 + 0.1209 = 0.1673

0.1673 = 16.73% probability that fewer than 3 weigh more than 20 pounds.

For more than 7, the probability is:

$P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 8) = C_{10,8}.(0.4)^{8}.(0.6)^{2} = 0.0106$

$P(X = 9) = C_{10,9}.(0.4)^{9}.(0.6)^{1} = 0.0016$

$P(X = 10) = C_{10,10}.(0.4)^{10}.(0.6)^{0} = 0.0001$

Since P(X > 7) < 0.05, it would be unusual if more than 7 of them weigh more than 20 pounds.

More can be learned about the binomial distribution at brainly.com/question/24863377

#SPJ1

4 0

1 year ago