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Ainat [17]
2 years ago
8

Mr. Torres took his students to the dolphin show. Each row in the stadium had 11 seats. One adult sat at each end of a row, and

each group of 4 students was seated between 2 adults. Mr. Torres sat by himself. How many adults were there?
Mathematics
1 answer:
Katena32 [7]2 years ago
3 0

Answer:

3

Step-by-step explanation:

The given data shows that only 1 row was in use. We also know that Mr. Torres is an adult. So for the ease, lets use A for Adult and S for student.

The seating must fulfill the requirements that 4 students must be between adults such that it would be "A-S-S-S-S-A"

On a row of 11 seats, this should be the searing arrangements.

A-S-S-S-S-A-S-S-S-S-A ( for 1 row).

Mr. Torres could be any A .

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The area is:
 A = x * y
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 P = 2x + 2y = 100
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 y = 50-x
 We write the area in terms of x:
 A (x) = x * (50-x)
 Rewriting:
 A (x) = 50x-x ^ 2
 Deriving:
 A '(x) = 50-2x
 We equal zero and clear x:
 50-2x = 0
 x = 50/2
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 Then, the other dimension is given by:
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8 0
2 years ago
WILL GIVE BRAINLIEST. Joanna saves nickels and quarters in a jar. If she saved five times as many nickels as quarters, and her c
vekshin1

Answer:

78 quarters

Step-by-step explanation:

I think because, I created an equation, 5x = 390 and solves to get 78

3 0
2 years ago
The subjects of a study by Dugoff et al. (A-5) were 10 obstetrics and gynecology interns at the University of Colorado Health Sc
astra-53 [7]

Answer:

The 95 percent confidence interval for the mean of the population from which the study subjects may be presumed to have been drawn is (19.1269, 32.6730).

Step-by-step explanation:

Intern            No. of Breast

Number        Exams Performed               X²

1                         30                                  900

2                        40                                 1600

3                        8                                        64

4                        20                                   400

5                          26                                 676

6                           35                               1225

7                            35                               1225

8                            20                                400

9                             25                              625

<u>10                                 20                        400 </u>

<u>                                   </u><u> ∑ 259                  ∑ 7515</u>

Mean= X`= ∑x/n= 259/10= 25.9

Variance = s²= 1/n-1[∑X²- (∑x)²/n]

= 1/0[7515- (259)²/10]= 1/9[7515- 6708.1]

= 806.9/9=89.655= 89.66

Standard Deviation= √89.655= 9.4687

Hence

The value of t with significance level alpha= 0.05 and 9 degrees of freedom  is t(0.025,9)= 2.262

The 95 % Confidence interval is given by

x`±t(∝,n-1) s/√n

So Putting the values

25.9± 2.262( 9.4687/√10)

= 25.9 ±2.262 (2.9943)

= 25.9 ± 6.7730

= 25.9 +6.7730=32.6730

25.9 -6.7730= 19.1269

= 19.1269, 32.6730

The 95 percent confidence interval for the mean of the population from which the study subjects may be presumed to have been drawn is (19.1269, 32.6730).

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