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evablogger [386]
3 years ago
13

A jogger is running around a 3/4 mile track and is 2/3 of the way around the track. How far has the jogger traveled

Mathematics
1 answer:
weeeeeb [17]3 years ago
3 0

Answer:

im literally doing the same exact thing right now

Step-by-step explanation:

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GIven f(x) = 3x + 2, Find f(x) = 8 (Solve for x)
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F(x)=8
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Divide 3 both sides
X=6/3
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Why do you think people who are good at math are usually successful on their jobs?
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3 years ago
Shelia's measured glucose level one hour after a sugary drink varies according to the normal distribution with μ = 117 mg/dl and
TiliK225 [7]

Answer:

The level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L is L = 127.1 mg/dl.

Step-by-step explanation:

To solve this problem, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are 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.

Central limit Theorem

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}}

In this problem, we have that:

\mu = 117, \sigma = 10.6, n = 6, s = \frac{10.6}{\sqrt{6}} = 4.33

What is the level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L ?

This is the value of X when Z has a pvalue of 1-0.01 = 0.99. So X when Z = 2.33.

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

By the Central Limit Theorem

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

2.33 = \frac{X - 117}{4.33}

X - 117 = 2.33*4.33

X = 127.1

The level L such that there is probability only 0.01 that the mean glucose level of 6 test results falls above L is L = 127.1 mg/dl.

6 0
3 years ago
Someone pls help asap this is due tmrw !!<br><br> x - 3y = -12<br> 2x + 3y = 3
hram777 [196]

Answer:

Step-by-step explanation:

to solve this system of equations we will add them

x-3y=-12

2x+3y=3 now +3y-3y=0

3x=-12+3

3x=-9

x=-3

plug in the first equation

-3-3y=-12

-3y=-12+3

-3y=-9

y=3

4 0
3 years ago
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