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3 years ago

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A process is turning out end items that have defects of type a, or type b, or both. if the probability of type a defect is 0.1 a

nd of type b is 0.2, the probability that an end item will have defects is:

1 answer:

Andreas93 [3]3 years ago

3 0

the probability will end up being
0.28

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Someone please help me answer this!!

lianna [129]

Answer:

7.1 units is the answer

Step-by-step explanation:

$d=\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \\\\d=\sqrt{(6-1)^2 + (2-7)^2}\\\\d=\sqrt{(5\\)^2+(-5)^2} \\\\d=\sqrt{25+25} \\\\d=\sqrt{50} \\\\d=7.071..$

Rounding off,

d = 7.1 units

3 0

3 years ago

Nine tiles are numbered $\color[rgb]{0.35,0.35,0.35}1, 2, 3, \ldots, 9$. Each of three players randomly selects and keeps three

Eduardwww [97]

The probability that all three players obtain an odd sum is 3/14.

<h3>What is probability?</h3>

The probability is the ratio of possible distributions to the total distributions.

I.e.,

Probability = (possible distributions)/(total distributions)

<h3>Calculation:</h3>

Given that,

There are nine tiles - 1, 2, 3,...9, respectively.

A player must have an odd number of odd tiles to get an odd sum. That means he can either have three odd tiles, or two even tiles and an odd tile.

In the given nine tiles the number of odd tiles = 5 and the number of even tiles = 4.

The only possibility is that one player gets 3 odd tiles and the other two players get 2 even tiles and 1 odd tile.

So,

One player can be selected in $^3C_1$ ways.

The 3 odd tiles out of 5 can be selected in $^5C_3$ ways.

The remaining 2 odd tiles can be selected and distributed in $^2C_1$ ways.

The remaining 4 even tiles can be equally distributed in $\frac{4 ! \cdot 2 !}{(2 !)^{2} \cdot 2 !}$ ways.

So, the possible distributions = $^3C_1$ × $^5C_3$ × $^2C_1$ × $\frac{4 ! \cdot 2 !}{(2 !)^{2} \cdot 2 !}$

⇒ 3 × 10 × 2 × 6 = 360

To find the total distributions,

The first player needs 3 tiles from the 9 tiles in $^9C3=84$ ways

The second player needs 3 tiles from the remaining 6 tiles in $^6C_3=20$ ways

The third player takes the remaining tiles in 1 way.

So, the total distributions = 84 × 20 × 1 = 1680

Therefore, the required probability = (possible distributions)/(total distributions)

⇒ Probability = 360/1680 = 3/14.

So, the required probability for the three players to obtain an odd sum is 3/14.

Learn more about the probability of distributions here:

brainly.com/question/2500166

#SPJ4

3 0

1 year ago

A manufacturer reported a sample mean = 22.0 g and a sample standard deviation = 2.5 g based on a sample of 20 of their products

tester [92]

Answer:

$n=(\frac{1.640(2.5)}{2})^2 =4.2 \approx 5$

So the answer for this case would be n=5 rounded up to the nearest integer

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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".

$\bar X=22$ represent the sample mean

$\mu$ population mean (variable of interest)

s=2.5 represent the sample standard deviation

n represent the sample size

ME = 2 represent the margin of error accepted

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (2)

And on this case we have that ME =2 and we are interested in order to find the value of n, if we solve n from equation (2) we got:

$n=(\frac{z_{\alpha/2} s}{ME})^2$ (3)

We can use as estimator for the population deviation the sample deviation $\hat \sigma = s$

The critical value for 90% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.05;0;1)", and we got $z_{\alpha/2}=1.640$ , replacing into formula (3) we got:

$n=(\frac{1.640(2.5)}{2})^2 =4.2 \approx 5$

So the answer for this case would be n=5 rounded up to the nearest integer

6 0

3 years ago

Find the volume of the composite figure. Explain your thought, explain how your arrived at your answer. SHOW YOUR WORK FOR FULL

Bond [772]

At the bottom of the composite figure we have half a sphere, of radius 10 in.

The volume of this hemisphere would be half the volume of the full sphere, or:

(1/2)(4/3)π(10 in)^3, or (2/3)π(1000 in^3), or (2000/3)π in^3.

On top is the cone of radius 10 and slant height 15 in. To find the volume of this cone-shaped solid, we'll need the height of the cone. This can be found using the Pyth. Thm. as follows:

15^2 = 10^2 + h^2, where h is the height of the cone.

225 = 100 + h^2, so that h= √125, or 5√5. The height of the cone is 5√5 in.

Then the volume of the cone is V = (1/3)(base)(height)

= (1/3)(π)(100 in^2)(5√5 in)

= 500√5/3(π) in^3

The total volume of the composite solid is then

(2000/3)(π in^3) + ( 500√5/3(π) ) in^3), or

(π/3)(4+√5) in^3. This comes out to 6.53 in^3, to the nearest hundredth.

7 0

3 years ago

Point Z on a coordinate grid is at (-1,3). If the point is reflected across the line y=-x, what are the coordinates of its image

Neko [114]

(-3,1)

x,y > -y,-x

Therefore (-1,3) is (-3,1)

4 0

4 years ago