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Svet_ta [14]
3 years ago
11

4. If(tanθ-cotθ)=4,then tan'0+cot0 is:​

Mathematics
1 answer:
Aneli [31]3 years ago
5 0
Jjsbsneakozhsbqmwkdldwmq
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Help me with this math question plz
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The range of crying times within 68% of the data  is (5.9, 8.1).

The range of crying times within 95% of the data  is (4.8, 9.2).

The range of crying times within 99.7% of the data  is (3.7, 10.3).

Step-by-step explanation:

According to the Empirical Rule in a normal distribution with mean µ and standard deviation σ, nearly all the data will fall within 3 standard deviations of the mean. The empirical rule can be broken into three parts:

  • 68% data falls within 1 standard deviation of the mean. That is P (µ - σ ≤ X ≤  µ + σ) = 0.68.
  • 95% data falls within 2 standard deviations of the mean. That is P (µ - 2σ ≤ X ≤  µ + 2σ) = 0.95.
  • 99.7% data falls within 3 standard deviations of the mean. That is P (µ - 3σ ≤ X ≤ µ + 3σ) = 0.997.

The mean and standard deviation are:

µ = 7

σ = 1.1

Compute the  range of crying times within 68% of the data as follows:

P(\mu-\sigma\leq X\leq \mu+\sigma)=0.68\\\\P(7-1.1\leq X\leq 7+1.1)=0.68\\\\P(5.9\leq X\leq 8.1)=0.68

The range of crying times within 68% of the data  is (5.9, 8.1).

Compute the  range of crying times within 95% of the data as follows:

P(\mu-2\sigma\leq X\leq \mu+2\sigma)=0.95\\\\P(7-2.2\leq X\leq 7+2.2)=0.95\\\\P(4.8\leq X\leq 9.2)=0.95

The range of crying times within 95% of the data  is (4.8, 9.2).

Compute the  range of crying times within 99.7% of the data as follows:

P(\mu-3\sigma\leq X\leq \mu+3\sigma)=0.997\\\\P(7-3.3\leq X\leq 7+3.3)=0.997\\\\P(3.7\leq X\leq 10.3)=0.997

The range of crying times within 99.7% of the data  is (3.7, 10.3).

8 0
3 years ago
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