Find the number of times a heart beats in 2 minutes if it beats 576 times in 8 minutes.

Can someone my answers

erik [133]

Answer:

Option 2 is correct that is $\frac{x+2}{x+9}$ and $x\neq-9$ and $x\neq-12$

Step-by-step explanation:

We have been given with the expression $\frac{x^2-10x-24}{x^2-3x-108}$

We will simplify the given expression by factorisation we will get $\frac{x^2-12x+2x-24}{x^2-12x+9x-108}$

Taking x as common from first two terms in numerator and 2 from last two terms in numerator

Similarly, take x common from first two terms and 9 from last two terms in denominator we will get

$\frac{x(x-12)+2(x-12)}{x(x-12)+9(x-12)}$

After arranging the terms we will get $\frac{(x+2)(x-12)}{(x+9)(x-12)}$

Taking out the common factor which is (x-12) from numerator and denominator it will get cancelled we will get

$\frac{(x+2)}{(x+9)}$

Hence, Option 2 is correct that is $\frac{x+2}{x+9}$ and $x\neq-9$ and $x\neq-12$

6 0

3 years ago

Given the tile pattern describe how the pattern is growing and find how many tiles will be in step seven

earnstyle [38]

Should be 13 Across the bottom and 6 on top which would be 19 in total

7 0

2 years ago

24% of 289= <br> Estimate using a rate per 100

Law Incorporation [45]

24% of 289 is 69.36 or 69 if you rounded. Hope it helps.

4 0

3 years ago

What is the solution to the system of equations? 2x+4t=12

katen-ka-za [31]

Answer:

C. $(8, -1)$

Step-by-step explanation:

{y = ¼x - 3}

{2x + 4y = 12

2x + 4[¼x - 3] = 12

2x + x - 12 = 12

3x - 12 = 12

+ 12 + 12

_________

3x = 24

__ __

3 3

$x = 8$ [Plug this back into both equations above to get the y-coordinate of −1]; $-1 = y$

I am joyous to assist you anytime.

6 0

3 years ago

The process standard deviation is 0.27, and the process control is set at plus or minus one standard deviation. Units with weigh

mr_godi [17]

Answer:

a) $P(X$

And for the other case:

tex] P(X>10.15)[/tex]

$P(X>10.15)= P(Z > \frac{10.15-10}{0.15}) = P(Z>1)=1-P(Z$

So then the probability of being defective P(D) is given by:

$P(D) = 0.159+0.159 = 0.318$

And the expected number of defective in a sample of 1000 units are:

$X= 0.318*1000= 318$

b) $P(X$

And for the other case:

tex] P(X>10.15)[/tex]

$P(X>10.15)= P(Z > \frac{10.15-10}{0.05}) = P(Z>3)=1-P(Z$

So then the probability of being defective P(D) is given by:

$P(D) = 0.00135+0.00135 = 0.0027$

And the expected number of defective in a sample of 1000 units are:

$X= 0.0027*1000= 2.7$

c) For this case the advantage is that we have less items that will be classified as defective

Step-by-step explanation:

Assuming this complete question: "Motorola used the normal distribution to determine the probability of defects and the number of defects expected in a production process. Assume a production process produces items with a mean weight of 10 ounces. Calculate the probability of a defect and the expected number of defects for a 1000-unit production run in the following situation.

Part a

The process standard deviation is .15, and the process control is set at plus or minus one standard deviation. Units with weights less than 9.85 or greater than 10.15 ounces will be classified as defects."

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

$X \sim N(10,0.15)$

Where $\mu=10$ and $\sigma=0.15$

We can calculate the probability of being defective like this:

$P(X$

And we can use the z score formula given by:

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

And if we replace we got:

$P(X$

And for the other case:

tex] P(X>10.15)[/tex]

$P(X>10.15)= P(Z > \frac{10.15-10}{0.15}) = P(Z>1)=1-P(Z$

So then the probability of being defective P(D) is given by:

$P(D) = 0.159+0.159 = 0.318$

And the expected number of defective in a sample of 1000 units are:

$X= 0.318*1000= 318$

Part b

Through process design improvements, the process standard deviation can be reduced to .05. Assume the process control remains the same, with weights less than 9.85 or greater than 10.15 ounces being classified as defects.

$P(X$

And for the other case:

tex] P(X>10.15)[/tex]

$P(X>10.15)= P(Z > \frac{10.15-10}{0.05}) = P(Z>3)=1-P(Z$

So then the probability of being defective P(D) is given by:

$P(D) = 0.00135+0.00135 = 0.0027$

And the expected number of defective in a sample of 1000 units are:

$X= 0.0027*1000= 2.7$

Part c What is the advantage of reducing process variation, thereby causing process control limits to be at a greater number of standard deviations from the mean?

For this case the advantage is that we have less items that will be classified as defective

5 0

3 years ago