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Sergeeva-Olga [200]
3 years ago
15

there are 14 girls trying out for volleyball. Each team will have six players. How many four teams will be made?

Mathematics
1 answer:
nikdorinn [45]3 years ago
3 0

Answer:

2

Step-by-step explanation:

14 divided by 6 = 2.33333333

only 6 in a group so you round it

and you get 2

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The perimeters of two similar triangles are 25 cm and 15 cm respectively. If one side of the first
Vanyuwa [196]

Answer:

5.4cm

Step-by-step explanation:

In similar triangles, the perimeter of the triangles will be in the ratio of their corresponding sides.

The ratio of their perimeter is  

perimeterofΔ2          15

---------------------   = -------------

perimeterofΔ1           25  

So, the side of triangle 2 will be  

  9  *  15                  15

----------------  =   --------------

​  sideofΔ2                25

=  

​⟹side of Δ2=           25  / 9∗15         =5.4.

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3 years ago
Please hurry algebra 2
Lelu [443]

Answer:

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Step-by-step explanation:

8 0
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A spinner has 10 equal sections numbered 1 through 10. What is the probability of
Nezavi [6.7K]

Answer:

The correct answer is C. 1/10.

Step-by-step explanation:

Given that a spinner has 10 equal sections numbered 1 through 10, to determine what is the probability of landing on the 6, the following calculation must be performed:

6 = 1 side only

10 = 10 options

1/10 = X

Therefore, the probability of the spinner landing on the 6 is 1/10.

3 0
3 years ago
What is the number 154 798 105 in expanded notation​
Rama09 [41]

Expanded form:

1 × 100,000,000
+ 5 × 10,000,000
+ 4 × 1,000,000
+ 7 × 100,000
+ 9 × 10,000
+ 8 × 1,000
+ 1 × 100
+ 0 × 10
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I hope it helps :)
4 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
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