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3 years ago

Is 2.82842712475 a rational number

Mathematics

1 answer:

Galina-37 [17]3 years ago

7 0

No---------------------------------

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At 9 AM you have run 2 miles. At 9:24 AM you have run 5 miles. What is your running rate in minutes per mile

rosijanka [135]

I think it's 8 b/c 24 minutes divided by 3 miles is 8 minutes per mile.

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3 years ago

In a shelter, the number of kennels is proportional to the number of dogs. The constant of proportionality in terms of dogs per

crimeas [40]

There are 9 kennels.

(# of dogs) = 3 (# of kennels)

27 = 3 (# of kennels)

27 / 3 = (# of kennels)

9 = (# of kennels)

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3 years ago

Given f (x) = 3x +4, find f (2). A 6 B) 7 C 10 D) 14.

Tom [10]

Answer:

F(2)= 10 or C

Step-by-step explanation:

You just need to plug the answers in.

F(x)=3x+4

f(2)=3(2)+4

f(2)=6+4

f(2)= 10

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3 years ago

In the figure shown, line AB is parallel to line CD. What is the

timurjin [86]

Since AB is a straight line, and straight lines are 180 degrees, you can add the two 62 degree angles.

62+62=124

180-124=56

angle x is 56 degrees

3 0

3 years ago

The scores of individual students on the American College Testing (ACT) Program College Entrance Exam have a normal distribution

tekilochka [14]

Answer:

The probability that the average of the scores of all 400 students exceeds 19.0 is larger than the probability that a single student has a score exceeding 19.0

Step-by-step explanation:

Xi~N(18.6, 6.0), n=400, Yi~Ber(p); Z~N(0, 1);

$P(0\leq X\leq 19.0)=P(\frac{0-\mu}{\sigma} \leq \frac{X-\mu}{\sigma}\leq \frac{19-\mu}{\sigma}), Z= \frac{X-\mu}{\sigma}, \mu=18.6, \sigma=6.0$

$P(-3.1\leq Z\leq 0.0667)=\Phi(0.0667)-\Phi (-3.1)=\Phi(0.0667)-(1-\Phi (3.1))=0.52790+0.99903-1=0.52693$

P(Xi≥19.0)=0.473

$\{Yi=0, Xi< 19\\Yi=1, Xi\geq 19\}$

p=0.473

Yi~Ber(0.473)

$P(\frac{1}{n}\displaystyle\sum_{i=1}^{n}X_i\geq 19)=P(\displaystyle\sum_{i=1}^{400}X_i\geq 7600)$

Based on the Central Limit Theorem:

$\displaystyle\sum_{i=1}^{n}X_i\~{}N(n\mu, \sqrt{n}\sigma),\displaystyle\sum_{i=1}^{400}X_i\~{}N(7440, 372)$

Then:

$P(\displaystyle\sum_{i=1}^{400}X_i\geq 7600)=1-P(0$

$P(\displaystyle\sum_{i=1}^{n}Y_i=1)=P(\displaystyle\sum_{i=1}^{400}Y_i=1)$

Based on the Central Limit Theorem:

$\displaystyle\sum_{i=1}^{400}Y_i\~{}N(400\times 0.473, \sqrt{400}\times 0.499)=\displaystyle\sum_{i=1}^{400}Y_i\~{}N(189.2; 9.98)$

$P(\displaystyle\sum_{i=1}^{400}Y_i=1)\~{=}P(0.5$

Then:

the probability that the average of the scores of all 400 students exceeds 19.0 is larger than the probability that a single student has a score exceeding 19.0

7 0

3 years ago