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quester [9]
4 years ago
7

Tracy started a job at $18,500. Each year her pay increases by $600. What is her total

Mathematics
1 answer:
Alex777 [14]4 years ago
3 0
24,500
600*10=6000
6000+18,500=24,500
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The ratio of the corresponding side lengths of two similar posters is 7 in:20 in. What is the ratio of the perimeters?
gregori [183]
The ratio of their perimeters is still 7:20. The poster is a rectangle shape. If the length of each side of a poster is proportional to the other by a certain ratio, the sum of the lengths of all sides is also proportional by the same ratio. You can suppose the lengths of four sides of poster A are 7a, 7b, 7c, 7d, so poster B has 20a, 20b, 20c, 20d. (7a+7b+7c+7d)/(20a+20b+20c+20d)=7(a+b+c+d)/20(a+b+c+d)=7/20.
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3 years ago
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Katen [24]
Which question are you asking ?
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4 years ago
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qaws [65]

Answer:

20

Step-by-step explanation:

(-4) × (-5) = 20

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3 years ago
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State whether the data described below are discrete or continuous, and explain why The durations of movies in minutes A Choose t
yKpoI14uk [10]

Answer:

D. The data are discrete because the data can only take on specific values.

Step-by-step explanation:

<u>Continuous variables</u> are variables that can take on any value in an interval. Examples of continuous variables are length, time, etc.

<u>Discrete variables</u> on the other hand are variables that can only take on specific values.

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4 0
3 years ago
suppose that incomes of families in Newport Harbor are normally distributed with a mean of $750,000 and a standard deviation of
denis23 [38]

Answer: 0.345

Step-by-step explanation:

Given : The incomes of families in Newport Harbor are normally distributed with Mean : \mu=\$ 750,000 and Standard deviation : \sigma= $250,000

Samples size : n=4

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The z-statistic :-

z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}

For x= $800,000

z=\dfrac{800000-750000}{\dfrac{250000}{\sqrt{4}}}=0.4

By using the standard normal distribution table , we have

The probability that the average income of these 4 families exceeds $800,000 :-

P(x>80,000)=P(z>0.4)=1-P(z\leq0.4)\\\\=1-0.6554217=0.3445783\approx0.345

Hence, the probability that the average income of these 4 families exceeds $800,000 =0.345

6 0
4 years ago
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