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3 years ago

What x-value makes the set of ratios equivalent? Tiles 2 : 3 = 6 : x ,4 : 7 = x : 42 , 2x : 48 = 3 : 12, 12 : 15 = x : 20

Mathematics

2 answers:

Sophie [7]3 years ago

6 0

the first and fifth

Naya [18.7K]3 years ago

4 0

Step-by-step explanation:

2 : 3 = 6 : x ,

Ratio can be written as

$\frac{2}{3} = \frac{6}{x}$

2*3 = 6, we multiply top and bottom of fraction by 3

$\frac{2*3}{3*3} = \frac{6}{9}$

So value of x=9

4 : 7 = x : 42 ,

$\frac{4}{7} = \frac{x}{42}$

7*6 = 42, we multiply top and bottom of fraction by 6

$\frac{4*6}{7*6} = \frac{24}{42}$

So value of x=24

2x : 48 = 3 : 12,

$\frac{2x}{48} = \frac{3}{12}$

\frac{3}{12}[/tex] = $\frac{2x}{48}12*4 = 48, we multiply top and bottom of fraction by 4[tex]\frac{3*4}{12*4} = \frac{12}{48}$

2x= 12, so x= 6

12 : 15 = x : 20

$\frac{12}{15} = \frac{x}{20}$

$15 * \frac{4}{3} = 20$ ,

we multiply top and bottom by 4/3

$12 * \frac{4}{3} = 16$

So value of x=16

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3 years ago

Genetic Defects Data indicate that a particular genetic defect occurs in of every children. The records of a medical clinic show

Dovator [93]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The probability that there exist 60 or more defected children is $P(x \ge 60)=0.0901$

Looking at the value for this probability we see that it is not so small to the point that the observation of this kind would be a rare occurrence

Step-by-step explanation:

From the question we are told that

in every 1000 children a particular genetic defect occurs to 1

The number of sample selected is $n= 50,000$

The probability of observing the defect is mathematically evaluated as

$p = \frac{1}{1000}$

$= 0.001$

The probability of not observing the defect is mathematically evaluated as

$q = 1-p$

$= 1-0.001$

$= 0.999$

The mean of this probability is mathematically represented as

$\mu = np$

Substituting values

$\mu = 50000*0.001$

$= 50$

The standard deviation of this probability is mathematically represented as

$\sigma = \sqrt{npq}$

Substituting values

$= \sqrt{50000 * 0.001 * 0.999}$

$= \sqrt{49.95}$

$= 7.07$

the probability of detecting $x \ge60$ defects can be represented in as normal distribution like

$P(x \ge 60)$

in standardizing the normal distribution the normal area used to approximate $P(x \ge 60)$ is the right of 59.5 instead of 60 because x= 60 is part of the observation

The z -score is obtained mathematically as

$z = \frac{x-\mu }{\sigma }$

$= \frac{59.5 - 50 }{7.07}$

$=1.34$

The area to the left of z = 1.35 on the standardized normal distribution curve is 0.9099 obtained from the z-table shown z value to the left of the standardized normal curve

Note: We are looking for the area to the right i.e 60 or more

The total area under the curve is 1

$P(x \ge 60) \approx P(z > 1.34)$

$= 1-P(z \le 1.34)$

$=1-0.9099$

$=0.0901$

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