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2 years ago

Plss help!Question: Find a,b and c

Mathematics

1 answer:

Roman55 [17]2 years ago

6 0

Answer:

See below, please

Step-by-step explanation:

The right：

${(y + c)}^{2} - b \times (x + a) \\ = ( {y}^{2} + 2yc + {c}^{2} ) - bx - ba =$

$= {y}^{2} - bx + 2cy + {c}^{2} - ab$

The left:

$= {y}^{2} - 7x - 18y + 109$

So, after comparing the right side and the left side of this equation, we know

$b = 7$

$2c = - 18 \\ so \\ c = - 9$

${c}^{2} - ab = 109$

So,

$a = \frac{ {c}^{2} - 109}{b} = \frac{ { (- 9)}^{2} - 109}{7} = \frac{ - 28}{7} = - 4$

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A public health organization reports that 40%of baby boys 6-8 months old in the United

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

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4 0

1 year ago

Need help please very confuzzled

SVEN [57.7K]

I'm not too sure either.

5(5) + 3(5) = 40 and

5(2) + 3 (10) = 40

If you can't think of anything maybe that'll help

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2 years ago

Plss help!Question: Find a,b and c​

Plss help!Question: Find a,b and c