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4 years ago

Math help, (again) These are mainly for corrections

Mathematics

2 answers:

dsp734 years ago

7 0

Answer:

Step-by-step explanation:

Because if you multiple 7 from 1,2,3,4 you can get all y answers like 7,14,21 and 28. HERE

7*1=7

7*2=14

7*3=21

7*4=28.

e-lub [12.9K]4 years ago

4 0

The answer would be the last option. :)

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Q varies inversely with (5-t). If Q = 8 when t = 3, find the value of: t when Q = 16

VMariaS [17]

Answer:

Step-by-step explanation:

Q*1/5-t

Q=k/5-t

8=k/5-3

8=k/2

8x2=16

k=16

Q=16/5-t

16=16/5-t

16(5-t)=16

80-16t=16

80-16=16t

64/16=t

t=4

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3 years ago

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4 years ago

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mina [271]

Answer:

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3 years ago

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Airida [17]

Answer:

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Step-by-step explanation:

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3 years ago

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Suppose that prices of a certain model of new homes are normally distributed with a mean of $150,000. use the 68-95-99.7 rule to

Novosadov [1.4K]

Answer:

We want to find the percentage of values between 147700 and 152300

$P(147700$

And one way to solve this is use a formula called z score in order to find the number of deviations from the mean for the limits given:

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

And replacing we got:

$z=\frac{147700-150000}{2300}=-1$

$z=\frac{152300-150000}{2300}=1$

So then we are within 1 deviation from the mean so then we can conclude that the percentage of values between $147,700 and $152,300 is 68%

Step-by-step explanation:

We define the random variable representing the prices of a certain model as X and the distirbution for this random variable is given by:

$X \sim N(\mu = 150000, \sigma =2300$

The empirical rule states that within one deviation from the mean we have 68% of the data, within 2 deviations from the mean we have 95% and within 3 deviations 99.7 % of the data.

We want to find the percentage of values between 147700 and 152300

$P(147700$

And one way to solve this is use a formula called z score in order to find the number of deviations from the mean for the limits given:

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

And replacing we got:

$z=\frac{147700-150000}{2300}=-1$

$z=\frac{152300-150000}{2300}=1$

So then we are within 1 deviation from the mean so then we can conclude that the percentage of values between $147,700 and $152,300 is 68%

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3 years ago