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Andrews [41]
3 years ago
13

a scale drawing of a rectangular park is 5 inches wide and 7 inches long. The actual park is 10,080 inches long What is the widt

h of the actual park, in inches
Mathematics
1 answer:
lilavasa [31]3 years ago
8 0

Shorter Answer:

7 inches → 280 yards

1 inch → 280 ÷ 7 = 40 yards

5 inches → 40 x 5 = 200 yards

Area = 280 x 200 = 56 000 yards

<em><u>Answer: 56 000 yards</u></em>

<em><u></u></em>

LongerAnswer (going more into depth):

<u><em>Area of the actual park is 56000 square yards.</em></u>

Step-by-step explanation:

A scale drawing of a rectangular park has dimensions as 5 inches width and 7 inches length.

Actual length of the park has been given as 280 yards.

So scale factor used in the drawing is 7 inches = 280 yards

Or 1 inch = 40 yards

Since width of the rectangular park is = 5 inch  

So actual width of the park will be = 5×40 = 200 yards

Now we can calculate the area of the park.

Actual area of the park = Length × Width

= 280×200

= 56000 yards²

<u><em>Therefore, area of the actual park is 56000 square yards.</em></u>

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The minimum score required for an A grade is 89.8.

Step-by-step explanation:

We are given that a psychology professor assigns letter grades on a test according to the following scheme. A : Top 7% of scores. B : Scores below the top 7% and above the bottom 64%. C : Scores below the top 36% and above the bottom 25%. D : Scores below the top 75% and above the bottom 6%. F : Bottom 6% of scores.

Scores on the test are normally distributed with a mean of 78.4 and a standard deviation of 7.6.

<u><em>Let X = Scores on the test</em></u>

SO, X ~ Normal(\mu=78.4,\sigma^{2} =7.6^{2})

The z-score probability distribution for normal distribution is given by;

                              Z = \frac{X-\mu}{\sigma} ~ N(0,1)

where, \mu = mean time = 78.4

            \sigma = standard deviation = 7.6

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.

Now, the minimum score required for an A grade so that it represents Top 7% of scores is given by;

      P(X \geq x) = 0.07   {where x is the required minimum score

      P( \frac{X-\mu}{\sigma} \geq \frac{x-78.4}{7.6} ) = 0.07

       P(Z \geq \frac{x-78.4}{7.6} ) = 0.07

<em>So, the critical value of x in the z table which represents the top 7% of the area is given as 1.4996, that is;</em>

                        \frac{x-78.4}{7.6} =1.4996

                        {x-78.4}{} =1.4996\times 7.6

                        x  = 78.4 + 11.39696 = <u>89.8 or 90</u>

Hence, the minimum score required for an A grade is 89.8.

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