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guapka [62]
3 years ago
8

Jim Tree wants to analyze his shipment of trees based on height. He knows the height of the trees is normally distributed so he

can use the standard normal distribution. He measures the height of 100 randomly selected trees in his shipment. He finds the mean is 65 inches and the standard deviation is 10 inches. What percentage of the trees will be between 55 inches and 75 inches?
Mathematics
2 answers:
Mazyrski [523]3 years ago
8 0
34%
hope this helps!!!
tester [92]3 years ago
3 0

Answer: If the mean is 65 inches, and the standard deviation is 10 inches, then when the problem ask for the percentage of the trees between 55 inches and 75 inches, is actually asking: which percentage of the trees is between the mean ± standard variation, or 65 inches ± 10 inches.

We know that if a distribution is normal, then the standard deviation encloses 34.1% of the set. But in this case, we are enclosing the area for  ± standard variation (this is two times the standard deviation, one for the left and one for the right) then you are enclosing 34.1% two times, and this is 68.2%.

A 68.2% of the trees is between 55 inches and 75 inches.

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Find the first four terms of the sequence an = 4n-6
natulia [17]

Answer:

The first four terms of the sequence are-6, -2, 2 and 6

Step-by-step explanation:

The given sequence is a_n=4n-6

In order to find the first four term of the sequence we put n =0, 1, 2 and 3.

For n =0

a_0=4(0)-6=-6

For n =1

a_1=4(1)-6=-2

For n =2

a_2=4(2)-6=2

For n =3

a_3=4(3)-6=6

Therefore, the first four terms of the sequence are

-6, -2, 2 and 6

5 0
3 years ago
Solve 47 (math operation)
Alchen [17]

Answer:

  t ≈ -2.014 or 3.647

Step-by-step explanation:

Add the opposite of the expression on the right side of the equal sign to put the equation into standard form.

  4.9t² -8t -36 = 0

You can divide by 4.9 to make this a little easier to solve.

  t² -(8/4.9)t -36/4.9 = 0

Now, add and subtract the square of half the x-coefficient to "complete the square."

  t² -(8/4.9)t +(4/4.9)² -36/4.9 -(4/4.9)² = 0

  (t -4/4.9)² -192.4/4.9² = 0 . . . . simplify

Add the constant term, then take the square root.

  (t -4/4.9)² = 192.4/4.9²

  t -4/4.9 = ±(√192.4)/4.9

  t = (4 ± √192.4)/4.9

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8 0
3 years ago
Given sin b = .88 find angle b in radians
vazorg [7]

first convert 0.88 to degrees

 so arcsin(0.88) = 61.642 degrees

 Arcsin on a calculator is sin^-1

then to convert to radians divide degrees by 180 and multiply that by PI

so (61.642/180) x PI = 1.076 radians

5 0
3 years ago
Kenyon ran for 8 miles each week while training. Here is his record of the number of miles he ran.
diamong [38]

Step-by-step explanation:

Let x be the miles Kenyon ran on tuesday

Let y be the miles Kenyon ran on thursday

Given,

Mon + Tues + Wed + Thu = 8mi

1.2 + x + 1.6 + y = 8 \\ 2.8 + x + y = 8 \\ x + y = 8 - 2.8 \\ x + y = 5.2

also given,

y = x + 0.4 \\  - x + y = 0.4

Now we have x + y = 5.2 as equation 1 and -x + y = 0.4 as equation 2

Using equation 1,

x + y = 5.2 \\ x = 5.2 - y

Substitute this equation into equation 2.

- (5.2 - y) + y = 0.4 \\  - 5.2 + y + y = 0.4 \\  - 5.2 + 2y = 0.4 \\ 2y = 0.4 + 5.2 \\ 2y = 5.6 \\ y = 5.6 \div 2 \\  = 2.8mi

Substitute y = 2.8 into equation 1.

x + 2.8 = 5.2 \\ x = 5.2 - 2.8 \\  = 2.4mi

Therefore Kenyon ran 2.4 miles on tuesday, and ran 2.8 miles on thursday.

8 0
2 years ago
Find the determinant of the nxn matrix A with 5's on the diagonal, l's above the diagonal, and O's below the diagonal. det (A) =
Vilka [71]

Answer:

n times 5

Step-by-step explanation:

A matrix Anxn of this way is called an upper triangular matrix. It can be proved that the determinant of this kind of matrix is  

a_{11}+a_{22}+...+a_{nn}

In this case, it would be 5+5+...+5 (n times) = n times 5

We are going to develop each determinant by the first column taking as pivot points the elements of the diagonal

det\left[\begin{array}{cccc}5&a_{12}&a_{13}...&a_{1n}\\0&5&a_{23}...&a_{2n}\\...&...&...&...\\0&0&0&5\end{array}\right] =5+det\left[\begin{array}{ccc}5&a_{23}...&a_{2n}\\0&5&a_{3n}\\...&...&...\\0&0&5\end{array}\right]=5+5+...+det\left[\begin{array}{cc}5&a_{n-1,n}\\0&5\end{array}\right]=5+5+...+5+5\;(n\;times)

6 0
3 years ago
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