Answer:
D
Step-by-step explanation:
The equation of a circle in standard form is
(x - h)² + (y - k)² = r²
where (h, k) are the coordinates of the centre and r is the radius
Here (h, k) = ( -
,
) and r =
, thus
(x - (-
))² + (y -
)² = (
)², that is
(x +
)² + (y -
)² =
→ D
<h2>Question #22 Answer</h2>
B. 2 in.
<h3>Explanation:</h3>
![\frac{3\times4}{6}](https://tex.z-dn.net/?f=%5Cfrac%7B3%5Ctimes4%7D%7B6%7D)
Cross out the common factor
![\frac{4}{2}\\ 2](https://tex.z-dn.net/?f=%5Cfrac%7B4%7D%7B2%7D%5C%5C%202)
________________________________________________________
<h2>Question #23 Answer (Picture attached)</h2>
D. proportional, equal
<h3>Explanation:</h3>
![Two\ figure\ are\ sand\ to\ \\ be\ similar\ if\ the\\ Conesesponding\ sides\ we\\ proportional\ and\ the\\ Corresponding\ ongles\ are\\ equal.](https://tex.z-dn.net/?f=Two%5C%20figure%5C%20are%5C%20sand%5C%20to%5C%20%5C%5C%20be%5C%20similar%5C%20if%5C%20the%5C%5C%20Conesesponding%5C%20sides%5C%20we%5C%5C%20proportional%5C%20and%5C%20the%5C%5C%20Corresponding%5C%20ongles%5C%20are%5C%5C%20equal.)
Δ![ABC\ is\ similar\ to](https://tex.z-dn.net/?f=ABC%5C%20is%5C%20similar%5C%20to)
Δ × ![yz.\ Thes](https://tex.z-dn.net/?f=yz.%5C%20Thes)
![\frac{AB}{xy}=\frac{AC}{xz}=\frac{BC}{yz}](https://tex.z-dn.net/?f=%5Cfrac%7BAB%7D%7Bxy%7D%3D%5Cfrac%7BAC%7D%7Bxz%7D%3D%5Cfrac%7BBC%7D%7Byz%7D)
![and](https://tex.z-dn.net/?f=and)
![< A= < X,\ < B= < y\ and\ < C= < z](https://tex.z-dn.net/?f=%3C%20A%3D%20%3C%20X%2C%5C%20%3C%20B%3D%20%3C%20y%5C%20and%5C%20%3C%20C%3D%20%3C%20z)
Answer:
7/25
Step-by-step explanation:
there are 14 numbers greater than 30 and less than 45
31 32 33 34 ......42 43 44 <=====14 numbers
14 choices out of 50 = 14/ 50 = 7/25
Answer:
A,B,D
Step-by-step explanation:
I just did it .
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)