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3 years ago

1/6 of 3 yards= --------feet is it 1 1/2 or 1/2 or 3 or 1/18

Mathematics

2 answers:

erica [24]3 years ago

8 0

1/2 yards times ) is 3, so 1/2 would be the answer.

Volgvan3 years ago

7 0

Answer:

1/2 ft or 6 inches

Step-by-step explanation:

1/6 of 3 yards = 1/2 ft or 6 inches

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if the local rate for electrical service is 6cents per kilowatt-hour how much money would be saved in a year by using a dishwash

dezoksy [38]

Given:

The local rate for electrical service is 6 cents per kilowatt-hour.

We need to know how much money would be saved in a year by using a dishwasher 4times per week rather than 6 times per week

From the table:

The cost of using a dishwasher 4 times per week = $29

The cost of using a dishwasher 6 times per week = $44

So, the saving will be = 44 - 29 = 15

So, the answer will be $15

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1 year ago

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2 years ago

4. Find the measure of each interior angle of a regular polygon with 12 sides.

professor190 [17]

For any polygon the sum of the exterior angles is 360. Since The polygon is regular divide 360 by 12 to get the exterior angle measure. Interior and exterior angles are supplemental so subtract the exterior angle measure from 180.

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3 years ago

Read 2 more answers

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

BRAINLIEST ASAP! PLEASE HELP ME :)

Phantasy [73]

See above explanation

I am joyous to assist you anytime.

7 0

3 years ago