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nekit [7.7K]
3 years ago
13

(10 POINTS AND BRAINLIEST FOR BEST). Last month Ben spent $380 on food and $420 on personal items.

Mathematics
1 answer:
lions [1.4K]3 years ago
6 0

4000 is his total monthly balance.

420 + 380 = 800

800/4000 = 0.20

0.20 = 20%

He spent 20% of his monthly income on food and personal items.

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H
Keith_Richards [23]

Answer:

3

Step-by-step explanation:

6 0
3 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
hoa [83]

Answer:

Obese people

Lower = \mu - 2\sigma = 373- 2(67) = 239

Upper = \mu + 2\sigma = 373+ 2(67) = 507

Lean People

Lower = \mu - 2\sigma = 526- 2(107) = 312

Upper = \mu + 2\sigma = 526+ 2(107) = 740

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".  

The empirical rule, also referred to as the three-sigma rule or 68-95-99.7 rule, is a statistical rule which states that for a normal distribution, "almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ). Broken down, the empirical rule shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ)".

Solution to the problem

Obese people

Let X the random variable that represent the minutes of a population (obese people), and for this case we know the distribution for X is given by:

X \sim N(373,67)  

Where \mu=373 and \sigma=67

On this case we know that 95% of the data values are within two deviation from the mean using the 68-95-99.7 rule so then we can find the limits liek this:

Lower = \mu - 2\sigma = 373- 2(67) = 239

Upper = \mu + 2\sigma = 373+ 2(67) = 507

Lean People

Let X the random variable that represent the minutes of a population (lean people), and for this case we know the distribution for X is given by:

X \sim N(526,107)  

Where \mu=526 and \sigma=107

On this case we know that 95% of the data values are within two deviation from the mean using the 68-95-99.7 rule so then we can find the limits liek this:

Lower = \mu - 2\sigma = 526- 2(107) = 312

Upper = \mu + 2\sigma = 526+ 2(107) = 740

The interval for the lean people is significantly higher than the interval for the obese people.

5 0
2 years ago
Somebody help me please this is algb 1
timama [110]

Answer:B

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
Does anyone know how to solve |4x+3|-9<5
elixir [45]

Answer in fraction form:   -17/4 < x < 11/4

Answer in decimal form:   -4.25 < x < 2.75

=======================================================

Work Shown:

|4x+3| - 9 < 5

|4x+3| < 5+9 .... adding 9 to both sides

|4x+3| < 14

-14 < 4x+3 <  14 .... see note at the bottom

-14-3 < 4x < 14-3 .... subtracting 3 from all sides

-17 < 4x < 11

-17/4 < x < 11/4 .... divide all sides by 4

-4.25 < x < 2.75

This says that x is between -4.25 and 2.75, but x is not equal to either endpoint.

Note: I used the rule that |x| < k becomes -k < x < k where k is a positive real number.

6 0
2 years ago
Delmar is constructing an equilateral triangle. He uses his straightedge to draw a segment and labels the endpoints W and X. Wit
MakcuM [25]
<span>Use a straightedge to join points W and P and then points P and X. â–łWPX is equilateral. Let's see now, Delmar has a line segment WX and has drawn 2 circles whose radius is the length of WX, centered upon W and centered upon X. Sounds to me that all he needs to do is select one of the intersections of those 2 circles and use that at the 3rd point of the equilateral triangle and draw a line from that point to W and another line from that point to X. Doesn't matter which of the two intersections he chooses, just needs to pick one. Looking at the available options, only the 1st one which is "Use a straightedge to join points W and P and then points P and X. â–łWPX is equilateral." matches my description, so that is the correct choice. The other choices tend to do rather bizarre things like create a perpendicular bisector of WX and for some unknown reason, claim that bisector is somehow a side of a desired equilateral triangle.</span>
7 0
2 years ago
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