Please help me out here.............

Solve for x 3x + 2 y;y=

Nataly [62]

Answer:

I beleive its 6

Step-by-step explanation:

5 0

4 years ago

Read 2 more answers

when your finding the area of a rectangle do you multiply the numbers around the perimeter or add them.

Flura [38]

When you are finding the area, you multiply the length x the width. So for example, we have a rectangle that the length is 6 feet and the width is 4 feet. So, first of all we multiply 6x4= 24. Since its area don't forget it is Square Feet (or any other type of measurement).

Hope it helps!

3 0

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

So

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

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

$=1-0.9099$

$=0.0901$

3 0

3 years ago

Pre-Calculus Help? Polar and Parametric Equations?

Paraphin [41]

$\bf x=rcos(\theta )\implies \cfrac{x}{cos(\theta )}=r\\\\
y=rsin(\theta )\implies \cfrac{y}{sin(\theta )}=r
\\\\
-------------------------------\\\\
r=3cos(\theta )\implies \cfrac{x}{cos(\theta )}=3cos(\theta )\implies x=3cos(\theta )cos(\theta )
\\\\\\
\boxed{x=3cos^2(\theta )}
\\\\\\
r=3cos(\theta )\implies \cfrac{y}{sin(\theta )}=3cos(\theta )\implies \boxed{y=3cos(\theta )sin(\theta )}$