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Yuliya22 [10]
3 years ago
13

(b) The area of a rectangular painting is 7719 cm2.

Mathematics
2 answers:
Nataliya [291]3 years ago
6 0

Given:

  • The area of a rectangular painting is 7719 cm²
  • Width of painting = 83cm

To Find?

  • Length of painting.

Solution:

  • Let length of painting be x cm

Using formula:

  • Area of rectangle = L × B

Where,

  • Length = x cm
  • Breadth = 73 cm

→ 7719 = x × 83

→ x = 7719/83

→ x = 93 cm

Hence,

  • Length of painting is 93cm
Salsk061 [2.6K]3 years ago
3 0

We are given that the width of of a rectangular painting is 83 cm and it's area is 7719 cm² and we are asked to find it's length , as we know that the area of a rectangular shape is given by <em><u>L×</u></em><em><u>B</u></em>. Where <em><u>L</u></em> is the length of the rectangular shape and <em><u>B</u></em> is the width/Breadth .Now , applying the same concept in our question we have ;

{:\implies \quad \sf L\times 83=7719}

{:\implies \quad \sf L=\dfrac{7719}{83}}

{:\implies \quad \sf L=93}

{:\implies \quad \bf \therefore \quad \underline{\underline{Length=93\:\: cm}}}

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6 0
3 years ago
The resting heart rate for an adult horse should average about µ = 47 beats per minute with a (95% of data) range from 19 to 75
KatRina [158]

Answer:

a. 0.0582 = 5.82% probability that the heart rate is less than 25 beats per minute.

b. 0.1762 = 17.62% probability that the heart rate is greater than 60 beats per minute.

c. 0.7656 = 76.56% probability that the heart rate is between 25 and 60 beats per minute

Step-by-step explanation:

Empirical Rule:

The Empirical Rule states that, for a normally distributed random variable:

Approximately 68% of the measures are within 1 standard deviation of the mean.

Approximately 95% of the measures are within 2 standard deviations of the mean.

Approximately 99.7% of the measures are within 3 standard deviations of the mean.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

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\mu = 47

(95% of data) range from 19 to 75 beats per minute.

This means that between 19 and 75, by the Empirical Rule, there are 4 standard deviations. So

4\sigma = 75 - 19

4\sigma = 56

\sigma = \frac{56}{4} = 14

a. What is the probability that the heart rate is less than 25 beats per minute?

This is the p-value of Z when X = 25. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{25 - 47}{14}

Z = -1.57

Z = -1.57 has a p-value of 0.0582.

0.0582 = 5.82% probability that the heart rate is less than 25 beats per minute.

b. What is the probability that the heart rate is greater than 60 beats per minute?

This is 1 subtracted by the p-value of Z when X = 60. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{60 - 47}{14}

Z = 0.93

Z = 0.93 has a p-value of 0.8238.

1 - 0.8238 = 0.1762

0.1762 = 17.62% probability that the heart rate is greater than 60 beats per minute.

c. What is the probability that the heart rate is between 25 and 60 beats per minute?

This is the p-value of Z when X = 60 subtracted by the p-value of Z when X = 25. From the previous two items, we have these two p-values. So

0.8238 - 0.0582 = 0.7656

0.7656 = 76.56% probability that the heart rate is between 25 and 60 beats per minute

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