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3 years ago

Ivision (Level 1) AM lt when 3x3 + 8x2 + 20x + 25 is divided by 3x + 5? Submit Answer

Mathematics

1 answer:

labwork [276]3 years ago

3 0

Answer: =3x^3+8x&2+ 85 /3 x+5

Step-by-step explanation:

3x3+8x2+20x+25 ÷ 3x+5

= 3x^3+8x^2+20x+25/3 x + 5

= (3x3)+(8x2)+(20x+ 25/3 x)+(5)

=3x^3+8x&2+ 85 /3 x+5

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2 years ago

The weight of people in a small town in Missouri is known to be normally distributed with a mean of 186 pounds and a standard de

OleMash [197]

Answer:

the probability that a random sample of 17 persons will exceed the weight limit of 3,417 pounds is 0.0166

Step-by-step explanation:

The summary of the given statistical data set are:

Sample Mean = 186

Standard deviation = 29

Maximum capacity 3,417 pounds or 17 persons.

sample size = 17

population mean =3417

The objective is to determine the probability that a random sample of 17 persons will exceed the weight limit of 3,417 pounds

In order to do that;

Let assume X to be the random variable that follows the normal distribution;

where;

Mean $\mu$ = 186 × 17 = 3162

Standard deviation = $29* \sqrt{17}$

Standard deviation = 119.57

$P(X>3417) = P(\dfrac{X - \mu}{\sigma}>\dfrac{X - \mu}{\sigma})$

$P(X>3417) = P(\dfrac{3417 - \mu}{\sigma}>\dfrac{3417 - 3162}{119.57})$

$P(X>3417) = P(Z>\dfrac{255}{119.57})$

$P(X>3417) = P(Z>2.133)$

$P(X>3417) =1- 0.9834$

$P(X>3417) =0.0166$

Therefore; the probability that a random sample of 17 persons will exceed the weight limit of 3,417 pounds is 0.0166

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3 years ago