Answer:
1: L = 5
W = 7
2: A = lw
3: 35
4: 18
5: You have to add them up because it is 1 shape broken up into different parts.
6: 53
Answer:
H is the same coordinates as G
The sentence that needs to be revised to improve the paragraph's sentence fluency is: sentence 3.
The two correct representations for the inequality can be -6x + 15 < 10 - 5x OR x < 5
In a grammatical context, the correct usage of punctuation marks and active voice is required and needed to be taken into consideration by the writer for effective communication to the reader.
The correct form of the sentence in sentence 3 should be:
- "Previously, only amateur players, like college athletes, were allowed to participate."
The inequality given is -3(2x - 5) < 5(2 - x). Using the simple algebraic method to solve the given inequality, we have:
= -3(2x - 5) < 5(2 - x)
= -6x + 15 < 10 - 5x
By collecting the like terms, we have:
= 15 - 10 < -5x + 6x
= x < 5
Therefore, the two correct representations for the inequality can be -6x + 15 < 10 - 5x OR x < 5
NOTE: The third question is a repetition of an incomplete question.
Learn more about sentences here:
brainly.com/question/4955765
Answer:
153.93804
Step-by-step explanation:
Area = π![r^2](https://tex.z-dn.net/?f=r%5E2)
r = ![\frac{d}{2}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7B2%7D)
Area = π
= 153.938040026
Answer:
a) The 90% confidence interval is ![0.575\leq \pi\leq 0.701](https://tex.z-dn.net/?f=0.575%5Cleq%20%5Cpi%5Cleq%200.701)
b) The margin of error is 0.063.
Step-by-step explanation:
We have to construct a 90% confidence interval for the proportion of students enrolled in college or a trade school within 12 months of graduating from high school in 2013.
We have a sample of 160 students, and a proportion of:
![p=X/N=102/160=0.6375](https://tex.z-dn.net/?f=p%3DX%2FN%3D102%2F160%3D0.6375)
The standard deviation is:
![\sigma=\sqrt{\frac{p(1-p)}{N}}=\sqrt{\frac{0.6375*0.3625}{160}}=0.038](https://tex.z-dn.net/?f=%5Csigma%3D%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7BN%7D%7D%3D%5Csqrt%7B%5Cfrac%7B0.6375%2A0.3625%7D%7B160%7D%7D%3D0.038)
As the sample size is big enough, we use the z-value as statistic. For a 90% CI, the z-value is z=1.645.
Then, the margin of error is:
![E=z\cdot\sigma=1.645\cdot 0.038=0.063](https://tex.z-dn.net/?f=E%3Dz%5Ccdot%5Csigma%3D1.645%5Ccdot%200.038%3D0.063)
Then, the 90% confidence interval is:
![p-z\cdot \sigma\leq \pi\leq p+z\cdot\sigma\\\\0.638-0.063\leq \pi\leq 0.638+0.063\\\\0.575\leq \pi\leq 0.701](https://tex.z-dn.net/?f=p-z%5Ccdot%20%5Csigma%5Cleq%20%5Cpi%5Cleq%20p%2Bz%5Ccdot%5Csigma%5C%5C%5C%5C0.638-0.063%5Cleq%20%5Cpi%5Cleq%200.638%2B0.063%5C%5C%5C%5C0.575%5Cleq%20%5Cpi%5Cleq%200.701)