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Natalka [10]
3 years ago
6

According to records in a large hospital, the birth weights of newborns have a symmetric and bell-shaped relative frequency dist

ribution with mean 6.8 pounds and standard deviation 0.5 Approximately what proportion of babies are born with birth weight under 6.3 pounds?
Mathematics
1 answer:
Kamila [148]3 years ago
8 0

Answer:

15.9% of babies are born with birth weight under 6.3 pounds.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 6.8 pounds

Standard Deviation, σ = 0.5

We are given that the distribution of  birth weights is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

P(birth weight under 6.3 pounds)

P(x < 6.3)

P( x < 6.3) = P( z < \displaystyle\frac{6.3 - 6.8}{0.5}) = P(z < -1)

Calculation the value from standard normal z table, we have,  

P(x < -1) = 0.159 = 15.9\%

15.9% of babies are born with birth weight under 6.3 pounds.

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kondaur [170]

The missing reason is (d) Add the fractions together on the right side of the equation

<h3>How to complete the missing reason?</h3>

From the statements, we have the following equation:

x^2 + b/a x + (b/2a)^2 = -4ac/4a^2 + b^2/4a^2

Next, we add the fractions on the right-hand side of the equation.

This gives

x^2 + b/a x + (b/2a)^2 = [-4ac + b^2]/4a^2

The above means that the last statement is gotten by adding the fractions on the right-hand side of the equation.

Hence, the missing reason is (d) Add the fractions together on the right side of the equation

Read more about quadratic equations at:

brainly.com/question/1214333

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8 0
2 years ago
find three consecutive odd integers such that the sum of the smallest number and middle number is 27 less than 3 times the large
katrin2010 [14]

A generic odd number can be written as

2k+1,\quad k \in \mathbb{Z}

Since there is an odd number every two numbers, three consecutive odd numbers will be

2k+1,\quad 2k+3,\quad 2k+5

Now let's make up the equations: the sum of the first two is

(2k+1)+(2k+3)

And 27 less than 3 times the largest is

3(2k+5)-27

These two must be the same, so we have

(2k+1)+(2k+3)=3(2k+5)-27 \iff 4k+4 = 6k+30-27 \iff 4k+4=6k+3

Subtracting 4k and 3 from both sides gives

1=2k \iff k=\dfrac{1}{2}

Which means that the problem has no solution.

To confirm this hypothesis, we can observe that, on the left hand side, we have the sum of two odd numbers, which is even

On the right hand side, we have an odd number, multiplied by 3 (still odd), take away 27 (still odd).

So, the left hand side is even, and the right hand side is odd. They can't be the same number.

7 0
3 years ago
An angle is equal to five times of its complement then what is its measures​
satela [25.4K]

Answer:

Then the complement of it = 90-x. Given that An angle is equal to 5 times its complement. x = 75. Hence the measure of an angle = 75.

Step-by-step explanation:

7 0
3 years ago
Find the area of the following polygon:
Masja [62]

Given:

ABCD is a quadrilateral

AC = 12, DE = 4 and FB = 5

To find:

The area of the polygon.

Solution:

AC bisects the quadrilateral into two triangles.

Area of triangle:

$A=\frac{1}{2}\times\text{base}\times\text{height}

<u>Area of triangle DAC:</u>

$A=\frac{1}{2}\times AC \times DE

$A=\frac{1}{2}\times 12 \times 4

A=24

Area of triangle DAC = 24 square units.

<u>Area of triangle BAC:</u>

$A=\frac{1}{2}\times AC \times FB

$A=\frac{1}{2}\times 12 \times 5

A=30

Area of triangle BAC = 30 square units.

Area of polygon = Area of triangle DAC + Area of triangle BAC

                           = 24 square units + 30 square units

                           = 54 square units

The area of the polygon is 54 square units.

3 0
3 years ago
4k – 6 = -2k – 16 – 2 for k
Contact [7]
<span>4k – 6 = -2k – 16 – 2
</span>4k+2k =  – 16 – 2+6
6k=-12
k=-2
Hope this helps
5 0
3 years ago
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