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8090 [49]
3 years ago
9

You are 55 feet tall and cast a 1010-foot shadow. A flagpole nearby casts a shadow that is 2828 feet. Which equation can you use

to solve for the height (h)h of the flagpole?
A. 10/5 = h/28

B. 5/10 = h/28

C. 5/28 = h/10

D. 5/10 = 28/h
Mathematics
1 answer:
sergejj [24]3 years ago
4 0
B. 5/10 = h/28
This is because 10 and 28 is simplified versions of 1010 and 2828, which are both the shadows, and they both either have to be on the top of the bottom, and 5 is a simplified version of 55 and 55 and 1010 have to go together
Hope this helps
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A rectangular table is 5 1/4 feet by 3 3/4 feet. What is the area of the table?
enot [183]

Answer:

19 11/16 or 315/16

Steps:

Turn the fractions into an improper fraction and then multiply straight across

5 1/4 = 21/4

3 3/4 = 15/4

(21/4)*(15/4)= 315/16= 19 11/16

Yes you got it right :)

4 0
3 years ago
Read 2 more answers
Eights rooks are placed randomly on a chess board. What is the probability that none of the rooks can capture any of the other r
erastova [34]

Answer:

The probability is \frac{56!}{64!}

Step-by-step explanation:

We can divide the amount of favourable cases by the total amount of cases.

The total amount of cases is the total amount of ways to put 8 rooks on a chessboard. Since a chessboard has 64 squares, this number is the combinatorial number of 64 with 8, 64 \choose 8 .

For a favourable case, you need one rook on each column, and for each column the correspondent rook should be in a diferent row than the rest of the rooks. A favourable case can be represented by a bijective function  f : A \rightarrow A , with A = {1,2,3,4,5,6,7,8}. f(i) = j represents that the rook located in the column i is located in the row j.

Thus, the total of favourable cases is equal to the total amount of bijective functions between a set of 8 elements. This amount is 8!, because we have 8 possibilities for the first column, 7 for the second one, 6 on the third one, and so on.

We can conclude that the probability for 8 rooks not being able to capture themselves is

\frac{8!}{64 \choose 8} = \frac{8!}{\frac{64!}{8!56!}} = \frac{56!}{64!}

7 0
2 years ago
Please help, must be done by today
Ostrovityanka [42]

The answer is -5x + 4y = -20

8 0
3 years ago
A sample of 20 joint specimens of a particular type gave a sample mean proportional limit stress of 8.41 mpa and a sample standa
nignag [31]

For this problem, the confidence interval is the one we are looking for. Since the confidence level is not given, we assume that it is 95%.

 

The formula for the confidence interval is: mean ± t (α/2)(n-1) * s √1 + 1/n


Where:

<span>
</span>

α= 5%


α/2 = 2.5%


t 0.025, 19 = 2.093 (check t table)


n = 20


df = n – 1 = 20 – 1 = 19

So plugging in our values:


8.41 ± 2.093 * 0.77 √ 1 + 1/20


= 8.41 ± 2.093 * 0.77 (1.0247)


= 8.41 ± 2.093 * 0.789019


= 8.41 ± 1.65141676


<span>= 6.7586 < x < 10.0614</span>

3 0
2 years ago
find the smallest number of terms of the AP "-54,-52.5,-51,-49.5" ....that must be taken for the sum of the terms to be positive
wel

The smallest number of terms of the AP that will make the sum of terms positive is 73.

Since we need to know the number for the sum of terms, we find the sum of terms of the AP

<h3>Sum of terms of an AP</h3>

The sum of terms of an AP is given by S = n/2[2a + (n - 1)d] where

  • n = number of terms,
  • a = first term and
  • d = common difference

Since we have the AP "-54,-52.5,-51,-49.5" ....", the first term, a = -54 and the second term, a₂ = -52.5.

The common difference, d = a₂ - a

= -52.5 - (-54)

= -52.5 + 54

= 1.5

<h3>Number of terms for the Sum of terms to be positive</h3>

Since we require the sum of terms , S to be positive for a given number of terms, n.

So, S ≥ 0

n/2[2a + (n - 1)d] ≥ 0

So, substituting the values of the variables into the equation, we have

n/2[2(-54) + (n - 1) × 1.5] ≥ 0

n/2[-108 + 1.5n - 1.5] ≥ 0

n/2[1.5n - 109.5] ≥ 0

n[1.5n - 109.5] ≥ 0

So, n ≥ 0 or 1.5n - 109.5 ≥ 0

n ≥ 0 or 1.5n ≥ 109.5

n ≥ 0 or n ≥ 109.5/1.5

n ≥ 0 or n ≥ 73

Since n > 0, the minimum value of n is 73.

So, the smallest number of terms of the AP that will make the sum of terms positive is 73.

Learn more about sum of terms of an AP here:

brainly.com/question/24579279

#SPJ1

4 0
2 years ago
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