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mr_godi [17]
3 years ago
9

Absolute value of 7 and one eigth

Mathematics
1 answer:
brilliants [131]3 years ago
3 0
The absolute value of 7 and 1/8 is 7 and 1/8! If you have to add these two values then it would be 7 1/8. Absolute value just means both values will be positive. Hope this helps!


Please give brainliest❤️
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Please help me
Nataly_w [17]

Jane is given a $100 gift to start and saves $35 a month from her allowance.

   After 1 month, Jane has saved

   After 2 months, Jane has saved

   After three months, Jane has saved

   and so on

In general, after x months Jane has saved

This means that it makes sense to represent the relationship between the amount saved and the number of months with one constant rate (in this case the constant rate is 35). It makes sense because the amount of money increases by $35 each month. Since the amount of increase is constant, we get constant rate. Also the initial amount is known ($100), so there is a possibility to write the equation of linear function representing this situation.

Step-by-step explanation:

5 0
3 years ago
Solve each proportion 3 = 72 – 7​
Vanyuwa [196]
1. 504=3x
x=168

2. 64=20m
m=3.2
4 0
2 years ago
It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minute
NemiM [27]

Answer:

a) 10.93% probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes.

b) 99.22% probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 minutes.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Mildly obese

Normally distributed with mean 373 minutes and standard deviation 67 minutes. So \mu = 373, \sigma = 67

A) What is the probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes?

So n = 5, s = \frac{67}{\sqrt{5}} = 29.96

This probability is 1 subtracted by the pvalue of Z when X = 410.

Z = \frac{X - \mu}{s}

Z = \frac{410 - 373}{29.96}

Z = 1.23

Z = 1.23 has a pvalue of 0.8907.

So there is a 1-0.8907 = 0.1093 = 10.93% probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes.

Lean

Normally distributed with mean 526 minutes and standard deviation 107 minutes. So \mu = 526, \sigma = 107

B) What is the probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 minutes?

So n = 5, s = \frac{107}{\sqrt{5}} = 47.86

This probability is 1 subtracted by the pvalue of Z when X = 410.

Z = \frac{X - \mu}{s}

Z = \frac{410 - 526}{47.86}

Z = -2.42

Z = -2.42 has a pvalue of 0.0078.

So there is a 1-0.0078 = 0.9922 = 99.22% probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 minutes.

7 0
3 years ago
The slope of any line perpendicular to AB , where A(0, -3) and B(4, 9) are two points on the line, is m =
rjkz [21]
Are you missing an option? The answer should be -2/3 (I just calculated it numerous times)
5 0
3 years ago
Algebra 1 IMPORTANT
babymother [125]

Answer:

A

Step-by-step explanation:

Subtract 1.6 from the height each hour and it shows that the answers match the height each hour.

5 0
3 years ago
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