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Jobisdone [24]
3 years ago
10

A value which represents any quantity along a number line

Mathematics
1 answer:
Ahat [919]3 years ago
6 0

Answer:

  C: real number

Step-by-step explanation:

The set of real numbers includes rational and irrational numbers. It would be difficult (impossible) to draw a number line consisting of only rational numbers or only irrational numbers.* So, the number lines we draw represent all real numbers.

_____

* There are an infinite number of rational numbers between any pair of irrational numbers. Similarly, there are an infinite number of irrational numbers between any pair of rational numbers

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Help me on number 12 13 14 and 15
aleksandr82 [10.1K]

12. 1.625 [terminating]; 13. 0.83 [bar notation over 3 (repeating)]; 14. 900 cm = 9 m; 15. 0.23 cm = 2.3 mm

Repeating decimals are parts of decimals that have repetitive digits; terminating decimals are decimals whose digits end.

Whether you are using Metric or Imperial, you have to determine whether you are going from a small unit to a big unit or vice versa. Then perform your operation. So, in exercise 14, the smaller unit is centimeters, so you would be going from big to small. Exercise 15 has you going from small to big.

There are centimeters in one meter, so multiply 9 by to get 900 centimeters.

There are 10 millimeters in one centimeter, so divide 2.3 by 10 simply by moving the decimal point ONCE to the left [Power of 10].

small to BIG → Division

BIG to small → Multiplication

I am joyous to assist you anytime.

5 0
3 years ago
Vail Resorts pays part-time seasonal employees at ski resorts on an hourly basis. At a certain mountain, the hourly rates have a
notsponge [240]

Answer:

z=0.842

And if we solve for a we got

\mu=13.16 +0.842*3=15.69

Step-by-step explanation:

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 hourly rates of a population, and for this case we know the distribution for X is given by:

X \sim N(\mu,3)  

For this part we want to find a value a, such that we satisfy this condition:

P(X>13.16)=0.20   (a)

P(X   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.80 of the area on the left and 0.20 of the area on the right it's z=0.842. On this case P(Z<0.842)=0.8 and P(z>0.842)=0.20

If we use condition (b) from previous we have this:

P(X  

P(z

But we know which value of z satisfy the previous equation so then we can do this:

z=0.842

And if we solve for a we got

\mu=13.16 +0.842*3=15.69

3 0
3 years ago
A programmer plans to develop a new software system. In planning for the operating system that he will​ use, he needs to estimat
Vedmedyk [2.9K]

Using the z-distribution, we have that:

a) A sample of 601 is needed.

b) A sample of 93 is needed.

c) A.  ​Yes, using the additional survey information from part​ (b) dramatically reduces the sample size.

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of \alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which z is the z-score that has a p-value of \frac{1+\alpha}{2}.

The margin of error is:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

95% confidence level, hence\alpha = 0.95, z is the value of Z that has a p-value of \frac{1+0.95}{2} = 0.975, so z = 1.96.

For this problem, we consider that we want it to be within 4%.

Item a:

  • The sample size is <u>n for which M = 0.04.</u>
  • There is no estimate, hence \pi = 0.5

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.04 = 1.96\sqrt{\frac{0.5(0.5)}{n}}

0.04\sqrt{n} = 1.96\sqrt{0.5(0.5)}

\sqrt{n} = \frac{1.96\sqrt{0.5(0.5)}}{0.04}

(\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.5(0.5)}}{0.04}\right)^2

n = 600.25

Rounding up:

A sample of 601 is needed.

Item b:

The estimate is \pi = 0.96, hence:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.04 = 1.96\sqrt{\frac{0.96(0.04)}{n}}

0.04\sqrt{n} = 1.96\sqrt{0.96(0.04)}

\sqrt{n} = \frac{1.96\sqrt{0.96(0.04)}}{0.04}

(\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.96(0.04)}}{0.04}\right)^2

n = 92.2

Rounding up:

A sample of 93 is needed.

Item c:

The closer the estimate is to \pi = 0.5, the larger the sample size needed, hence, the correct option is A.

For more on the z-distribution, you can check brainly.com/question/25404151  

8 0
2 years ago
Need help!!!
Assoli18 [71]

Answer:

73 mph during the second hour!

Step-by-step explanation:

Hour 1=54 mph

Hour 2=? mph

Hour 3=82 mph

Total=209 mph

this means that 209=82+54+?

Simplify it and you will get 73

73 mph

7 0
3 years ago
Read 2 more answers
Please hurry ;^;
kompoz [17]

Answer:

28x+8y+2

Step-by-step explanation:

the answer is 28x+8y+2 because a math calculator said so (m a t h w a y)

7 0
3 years ago
Read 2 more answers
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