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zhuklara [117]
2 years ago
6

24 mi per min = yards per min

Mathematics
1 answer:
slavikrds [6]2 years ago
3 0
42240 yards / minute
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The complementary angle of 45
Licemer1 [7]

Answer:

45°

Step-by-step explanation:

Complementary angles sum to 90°, thus

90° - 45° = 45° ← is the complement of 45°

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3 years ago
Margo’s Dad offers to give her 5 cents on sunday. Then for each day of the week, he offers to give her 5 times the amount from t
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Step-by-step explanation:

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3 years ago
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The distribution of weights for newborn babies is approximately normally distributed with a mean of 7.4 pounds and a standard de
blsea [12.9K]

Answer:

1. 15.87%

2.  6 pounds and 8.8 pounds.

3. 2.28%

4. 50% of newborn babies weigh more than 7.4 pounds.

5. 84%

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 7.4 pounds

Standard Deviation, σ = 0.7 pounds

We are given that the distribution of weights for newborn babies is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

1.Percent of newborn babies weigh more than 8.1 pounds

P(x > 8.1)

P( x > 8.1) = P( z > \displaystyle\frac{8.1 - 7.4}{0.7}) = P(z > 1)

= 1 - P(z \leq 1)

Calculation the value from standard normal z table, we have,  

P(x > 8.1) = 1 - 0.8413 = 0.1587 = 15.87\%

15.87% of newborn babies weigh more than 8.1 pounds.

2.The middle 95% of newborn babies weight

Empirical Formula:

  • Almost all the data lies within three standard deviation from the mean for a normally distributed data.
  • About 68% of data lies within one standard deviation from the mean.
  • About 95% of data lies within two standard deviations of the mean.
  • About 99.7% of data lies within three standard deviation of the mean.

Thus, from empirical formula 95% of newborn babies will lie between

\mu-2\sigma= 7.4-2(0.7) = 6\\\mu+2\sigma= 7.4+2(0.7)=8.8

95% of newborn babies will lie between 6 pounds and 8.8 pounds.

3. Percent of newborn babies weigh less than 6 pounds

P(x < 6)

P( x < 6) = P( z > \displaystyle\frac{6 - 7.4}{0.7}) = P(z < -2)

Calculation the value from standard normal z table, we have,  

P(x < 6) =0.0228 = 2.28\%

2.28% of newborn babies weigh less than 6 pounds.

4. 50% of newborn babies weigh more than pounds.

The normal distribution is symmetrical about mean. That is the mean value divide the data in exactly two parts.

Thus, approximately 50% of newborn babies weigh more than 7.4 pounds.

5. Percent of newborn babies weigh between 6.7 and 9.5 pounds

P(6.7 \leq x \leq 9.5)\\\\ = P(\displaystyle\frac{6.7 - 7.4}{0.7} \leq z \leq \displaystyle\frac{9.5-7.4}{0.7})\\\\ = P(-1 \leq z \leq 3)\\\\= P(z \leq 3) - P(z < -1)\\= 0.9987 -0.1587= 0.84 = 84\%

84% of newborn babies weigh between 6.7 and 9.5 pounds.

7 0
3 years ago
1. Which of the following is equivalent to the
tamaranim1 [39]

Answer: D.

Step-by-step explanation: Apply the distributive property.

4

x

+

4

⋅

3

Multiply  

4

 by  

3

.

4

x

+

12

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Find all solutions of the equation x^2 -2 x + 6 = 0 by completing the square. Express the soutions in the form a + b i:
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3 years ago
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