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3 years ago

The length of the path of the first swing of the bob of a pendulum is 0.6 in. Each

Mathematics

1 answer:

ycow [4]3 years ago

8 0

Answer:

0.3188646

Step-by-step explanation:

an=a×r^n-1

a6=0.6×0.9^5

=0.354294

1st swing: 0.6

2nd swing: 0.9×0.6=0.54

3rd swing: 0.9×0.54= 0.486

4th 0.9×0.486= 0.4374

5th 0.9×0.4374= 0.39366

6th 0.9×0.39366=0.354294

Brainliest please~

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Select the number line that shows that two opposite numbers have a sum of 0

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Answer:

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2 years ago

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2 years ago

Suppose a student's score on a standardize test to be a continuous random variable whose distribution follows the Normal curve.

blsea [12.9K]

Answer:

a) Percentage of students scored below 300 is 1.79%.

b) Score puts someone in the 90th percentile is 638.

Step-by-step explanation:

Given : Suppose a student's score on a standardize test to be a continuous random variable whose distribution follows the Normal curve.

(a) If the average test score is 510 with a standard deviation of 100 points.

To find : What percentage of students scored below 300 ?

Solution :

Mean $\mu=510$ ,

Standard deviation $\sigma=100$

Sample mean $x=300$

Percentage of students scored below 300 is given by,

$P(Z\leq \frac{x-\mu}{\sigma})\times 100$

$=P(Z\leq \frac{300-510}{100})\times 100$

$=P(Z\leq \frac{-210}{100})\times 100$

$=P(Z\leq-2.1)\times 100$

$=0.0179\times 100$

$=1.79\%$

Percentage of students scored below 300 is 1.79%.

(b) What score puts someone in the 90th percentile?

90th percentile is such that,

$P(x\leq t)=0.90$

Now, $P(\frac{x-\mu}{\sigma} < \frac{t-\mu}{\sigma})=0.90$

$P(Z< \frac{t-\mu}{\sigma})=0.90$

$\frac{t-\mu}{\sigma}=1.28$

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$t-510=128$

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3 years ago

Analyze the diagram below and complete the instructions that follow.

Trava [24]

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$\cos B = \dfrac{16^2 + 16^2 - 11^2}{2(16)(16)} = \dfrac{ 391}{512}$

$B = \arccos \dfrac{ 391}{512} \approx 40.2 ^\circ$

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3 years ago

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