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Illusion [34]
4 years ago
9

A​ city's mean minimum daily temperature in February is 22 degrees Upper F. Suppose the standard deviation of the minimum temper

ature in February is 6 degrees Upper F and that the distribution of minimum temperatures in February is approximately Normal. What percentage of days in February has minimum temperatures below freezing left parenthesis 32 degrees Upper F right parenthesis​?
Mathematics
1 answer:
Misha Larkins [42]4 years ago
7 0

Answer:

Probability that X <32 degrees

=P(Z

Convert this into percent by multiplying by 100

95.22% of days in February has minimum temperatures below freezing (32 degrees Upper F )​

Step-by-step explanation:

Given that a city's mean minimum daily temperature in February is 22 degrees Upper F, with std deviation 6 degrees.

Let X be the minimum temperature in February

Then X is N(22,6)

To calculate  percentage of days in February has minimum temperatures below freezing (32 degrees Upper F )

Probability that X <32 degrees

=P(Z

Convert this into percent by multiplying by 100

95.22% of days in February has minimum temperatures below freezing (32 degrees Upper F )​

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7 x 1 = 7 shows the _______ property
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multiplication property of one or identity

Step-by-step explanation:

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2 years ago
How many pairs of diagonals of a regular decagon intersect?
mars1129 [50]
I think it is 2 pairs
3 0
4 years ago
In a set of three lines, a pair of parallel lines is intersected by a transversal. The intersection forms eight special angles.
My name is Ann [436]

Answer:

Part A: 4x = -5(x-18)

Part B: A and B both = 40 degrees

Step-by-step explanation:

Ok, corresponding angles are in the same position on the two parallel lines cut by the transversal. And they are equal in measure. So, upper right = upper right.

So A = B

4x = -5(x -18)

4x = -5x + 90

9x = 90

x = 10

Angle A = 4x = 40 degrees, which means angle B is also 40, but let's show the work:

Angle B = -5(x - 18) = -50 + 90 = 40

7 0
3 years ago
Find a formula for the described function. A rectangle has perimeter 8 m. Express the area A of the rectangle as a function of t
Maksim231197 [3]

Answer:

A(L) = 4L - L^2

Step-by-step explanation:

Given

Perimeter = 8m

Required

Determine its area as a function of length

Represent Length and Width with L and W, respectively;

Perimeter (P) is calculated as thus;

P = 2(L + W)

Substitute 8 for P

8 = 2(L + W)

Divide both sides by 2

4 = L + W

Make W the subject of formula

W = 4 - L

Area (A) of a rectangle is calculated as thus:

A = L * W

Substitute 4 - L for W

A = L * (4 - L)

Open bracket

A = 4L - L^2

Represent as a function

A(L) = 4L - L^2

6 0
3 years ago
Which of the given numbers is not a Pythagorean triplets. *
Ludmilka [50]

Step-by-step explanation:

By Pythagoras' Theorem,

{c}^{2}  =  {a}^{2}  +  {b}^{2}

where c is always the largest number.

a and b can be interchangeable between the 2nd largest and the 3rd largest numbers.

Given a = 8, b = 15 and c = 17,

{a}^{2}  + b {}^{2}  =  {8}^{2}  +  {15}^{2}  \\  = 64 +  225 \\  = 289 \\   \\  {c}^{2}  =  {17}^{2}  \\  = 289

Since c^2 = a^2 + b^2 , 8 , 15 and 17 are pythagorean triplets.

Now let's move on to 9, 40 and 41.

{a}^{2}  +  {b}^{2}  =  {9}^{2}   +  {40}^{2}  \\  = 81 + 1600 \\  = 1681 \\  \\  {c}^{2}  =  {41}^{2}  \\  = 1681

Since c^2 = a^2 + b^2 , 9 , 40 and 41 are pythagorean triplets.

Last let's move on to 4,7 and 8.

{a}^{2}  +  {b}^{2}  =  {4}^{2}  +  {7}^{2}  \\  = 16 + 49 \\  = 65 \\  \\  {c}^{2}  =  {8}^{2}  \\  = 64

Since a^2+b^2 IS NOT EQUAL to c^2, 4,7 and 8 ARE NOT pythagorean triplets.

8 0
3 years ago
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