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NeX [460]
2 years ago
7

Marks on a public health test follow a normal distribution with a mean of 77 and a standard deviation of 11. What is the approxi

mate 40th percentile of the mark distribution (40% of data is equal or less than what mark)?
Mathematics
1 answer:
Kazeer [188]2 years ago
8 0

Answer:

58.

Step-by-step explanation:

We have been given that marks on a public health test follow a normal distribution with a mean of 77 and a standard deviation of 11. We are asked to find the approximate 40th percentile of the mark distribution.

We will use z-score formula and normal distribution table to solve our given problem.

z=\frac{x-\mu}{\sigma}, where,

z = Z-score,

x = Sample score,

\mu = Mean,

\sigma = Standard deviation.

z=\frac{x-77}{11}

From normal distribution table, we need to find z-score corresponding to 40th percentile or 0.40.

-1.75=\frac{x-77}{11}

Let us solve for x.

-1.75*11=\frac{x-77}{11}*11

-19.25=x-77

-19.25+77=x-77+77

57.75=x

x\approx 58

Therefore, the approximate 40th percentile of the mark distribution would be 58.

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Dimas [21]

Answer:

The ratio of male nurses to female nurse in lowest terms is: 2:5

Step-by-step explanation:

Given: In a hospital, there are  total 250 registered nurses.

Number of nurses are males = 100

Number of nurses are females = 250 - 100 = 150  [ Total nurses - number of male nurses]

To find the ratio of male nurse to female nurses in lowest terms;

A ratio shows that the relative sizes of two or more values.

Then,

\frac{Number of male nurses}{Number of female nurses} = \frac{100}{250}

or

= \frac{100}{250} =\frac{10}{25} =\frac{2}{5} = 2:5

Therefore, the ratio of male nurses to female nurses in lowest terms is; 2 :5

7 0
2 years ago
Read 2 more answers
Find dy/dx implicitly and find the largest interval of the form −a < y < a or 0 < y < a such that y is a differentia
mylen [45]

Answer:

dy/dx = -1/√(1 - x²)

For 0 < y < π

Step-by-step explanation:

Given the function cos y = x

-siny dy = dx

-siny dy/dx = 1

dy/dx = -1/siny (equation 1)

But cos²y + sin²y = 1

=> sin²y = 1 - cos²y

=> siny = √(1 - cos²y) (equation 2)

Again, we know that

cosy = x

=> cos²y = x² (equation 3)

Using (equation 3) in (equation 2), we have

siny = √(1 - x²) (equation 4)

Finally, using (equation 4) in (equation 1), we have

dy/dx = -1/√(1 - x²)

The largest interval is when

√(1 - x²) = 0

=> 1 - x² = 0

=> x² = 1

=> x = ±1

So, the interval is

-1 < x < 1

arccos(1) < y < arxcos(-1)

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6 0
2 years ago
Find the area of quadrilateral ABCD. Round the area to the nearest whole number, if necessary. A(-5, 4) 4 B(0, 3) 2. F(-2, 1) -2
valentina_108 [34]

Answer:26

Step-by-step explanation:

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2 years ago
A rectangular piece of land is 40m long and 25m wide. A path of uniform width and 426 m^2 area sorrounds
fredd [130]

Answer:

Width of the uniform path that surrounds the piece of land = 3 m

Step-by-step explanation:

Let the width of the path that surrounds the piece of land be x.

The situation described is sketched in the attached image to this solution.

The dimension of the Length of the piece of land including the uniform path that surrounds it = (40 + 2x) m

The Breadth of the piece of land including the uniform path that surrounds it = (25 + 2x) m

The area of the piece of land including the uniform path that surrounds it

= (Area of the piece of land) + (Area of the uniform path that surrounds it)

Area of the piece of land = Length × Breadth = 40 × 25 = 1000 m²

Area of the uniform path that surrounds it = 426 m² (Given in the question)

The area of the piece of land including the uniform path that surrounds it = 1000 + 426 = 1426 m²

But the area of the piece of land including the uniform path that surrounds it is also

= (Length of the piece of land including the uniform path that surrounds it) × (Breadth of the piece of land including the uniform path that surrounds it

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= 1000 + 80x + 50x + 4x²

= (4x² + 130x + 1000) m²

Equation these 2 areas

4x² + 130x + 1000 = 1426

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Solving the quadratic equation

x = 3 or -35.5

Since the width cannot be negative,

x = 3 m

Hope this Helps!!!

7 0
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Westkost [7]

We are given the equation:

<span>Z(q) = 4q + ½    ---> 1</span>

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<span>q = u + ½          ---> 2</span>

Substituting this value into equation # 1:

Z = 4 (u + ½) + ½ = z (u + 1/2) 

4 u + 2 + ½ = z (u + 1/2) 

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4 u + 2 + ½ = ½

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4 u = -2

u = -1/2         (ANSWER)

 

 

<span> </span>

5 0
2 years ago
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