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Nat2105 [25]
2 years ago
5

A farmer has 350 feet of fence available to enclose a 6125 square foot region in the shape of adjoining squares with sides of le

ngth x and y. The big square has sides of x and the small square has sides of length y. Find x and y

Mathematics
1 answer:
denis-greek [22]2 years ago
4 0

Answer:

  x = 77 ft; y = 14 ft

Step-by-step explanation:

We assume that "adjoining squares" means the small square shares a side with the large square, so the total length (in feet) of fence required for the enclosure is ...

  4x +3y = 350

The sum of the two areas is 6125 ft², so we have another relation:

  x² +y² = 6125

We can use the first equation to write an expression for y, then substitute that into the second equation. The result is a quadratic in x.

  3y = 350 -4x

  y = (350 -4x)/3

Then ...

  x² + ((350 -4x)/3)² = 6125 . . . substitute for y

  9x² +(350 -4x)² = 55125 . . . . multiply by 9

  25x² -2800x +67375 = 0 . . . . subtract 55125 and simplify

  x² -112x +2695 = 0

  (x -35)(x -77) = 0

This has two solutions: x = 35 and x = 77. We know that the square of sides x must use more than half the fence, so it must have side lengths greater than ...

  (350/2)/4 = 43.75 . . . . feet

The appropriate choice for x is 77 feet. Then ...

  y = (350 -4x)/3 = 42/3 = 14 . . . feet

x and y are 77 feet and 14 feet, respectively.

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One way to measure whether the trees in the Wade Tract are uniformly distributed is to examine the average location in the north
tatyana61 [14]

Answer:

Null hypothesis be H₀ : μ = 100

Alternative hypothesis Hₐ : μ < 100

There is sufficient statistical evidence to suggest that the average location of Wade Tract is 100

Step-by-step explanation:

Here we have;

Let our null hypothesis be H₀ : μ = 100 Average location of trees in the Wade Tract is 100

Our alternative hypothesis becomes Hₐ : μ < 100 at 95% confidence level

Proposed average location in Wade Tract, μ = 100

Sample mean, \bar x = 99.74

Standard deviation, s = 58

Sample size, n = 584

The t test formula is therefore;

t=\frac{\bar{x}-\mu }{\frac{s }{\sqrt{n}}}

Therefore, with df = 584 -1 = 583, and α = (1 - 0.95)/2 = 0.025

We have t_{\alpha /2} = -1.65

Plugging in the values into the t test formula, we have t = -0.108338, from which the p-value is given as p = 0.4569 which is much more than the value of α, therefore, we accept the null hypothesis as follows;

There is sufficient statistical evidence to suggest that the average location of Wade Tract = 100.

7 0
2 years ago
Clint is in study hall reading the hunger games after 20 minutes he is on page 151 after 45 minutes he's on page 181 how many pa
Mademuasel [1]

After 20 minutes Clint is on page = 151

After 45 minutes Clint is on page = 181

Time slot = 45-20=25 minutes

Difference in pages in 25 minutes = 181-151=30 pages.

Lets suppose Clint is reading each page at same pace.

So , In 25 minutes, he can read = 30 pages

In 1 minute he can read = \frac{30}{25}=1.2 pages

Initial time is 20 minutes, so he can read the number of pages in 20 minutes = 20\times1.2=24

So, number of pages Clint read prior to study hall were =

151-24=127 pages.

6 0
2 years ago
A gene can be either type A or type B, and it can be either dominant or recessive. If the gene is type B, then there is a probab
WARRIOR [948]

Answer: The probability that a gene is of type A is 0.29.

Step-by-step explanation:

Let A be the event of type A.

Let B be the event of type B.

Let D be the event of dominant.

So, P(B and D) = 0.22

P(D|B)=0.31

So, Using the conditional probability, we get that

P(B|D)=\dfrac{P(B\cap D)}{P(B)}\\\\0.31=\dfrac{0.22}{P(B)}\\\\P(B)=\dfrac{0.22}{0.31}=0.71

So, the probability that a gene is of type A is given by

P(A)=1-P(B)=1-0.71=0.29

Hence, the probability that a gene is of type A is 0.29.

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Round 14 17/21 to the nearest whole number
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Answer:

15 Is Your Answer

Step-by-step explanation:

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Hope you can see the picture

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