Refer to OP for Exercises 1-8. If SN and MT are diameters with m/ SPT 51 and m/ NPR 29, determine whether each arc is a minor ar
c, a major arc, or a semicircle. Then find the degree measure of each arc.
1. MNR
3. mTSR
2. MST
4. mMST
If MT= 15, find the length of each arc. Round to the nearest tenth.
5. NR
7. TSR
6. ST
8. MST
M
N
R
Geometry
1. MNR
3. mTSR
2. MST
4. mMST
If MT= 15, find the length of each arc. Round to the nearest tenth.
5. NR
7. TSR
6. ST
8. MST
M
N
R
Geometry
1 answer:
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Answer:
Step-by-step explanation:
Adding both equations will make it:
Putting x = 10 will make it:
= 3(1000)-40+1
= 3000-39
= 2961
Answer:
D. No, because the sample size is large enough.
Step-by-step explanation:
The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
If the sample size is higher than 30, on this case the answer would be:
D. No, because the sample size is large enough.
And the reason is given by The Central Limit Theorem since states if the individual distribution is normal then the sampling distribution for the sample mean is also normal.
From the central limit theorem we know that the distribution for the sample mean is given by:
If the sample size it's not large enough n<30, on that case the distribution would be not normal.