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OlgaM077 [116]
3 years ago
9

What is the value of the underlined digits? 588_7, 6.409_7, 8.41_3, _0,184,843

Mathematics
2 answers:
Georgia [21]3 years ago
8 0
First one: ten
second one: millionth place
third one: thousandth place
fourth one: million
irga5000 [103]3 years ago
3 0
The first on is the tenth place
the second one is  the tenth place
the third one is the tenth place
the last one is the millionth place
 
hope i helped :) i might be wrong though


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I am so done please help !!
ycow [4]

Answer: 24.2

Step-by-step explanation:

8 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
Holly has 38 pecans. She has 5 friends coming to visit. If Holly and her five friends each eat an equal number of pecans, about
MakcuM [25]

Answer: The answer is 6.3

5 0
3 years ago
How to do this question plz answer my question step by step plzz plz plz plz ​
never [62]

Answer:

40°, 95°, 105°, and 160°

Step-by-step explanation:

Let the smallest angle be x.

Measures of the 3 angles can be expressed as:

x + 55

x + 65

x + 120

The sum of all angles in a quadrilateral = 360°.

Therefore, (x + 55) + (x + 65) + (x + 120) = 360

Solve for x

x + 55 + x + 65 + x + 120 = 360

3x + 240 = 360

3x + 240 - 240 = 360 - 240

3x = 120

\frac{3x}{3} = \frac{120}{3}

x = 40

The smallest angle = 40°

Plug in the value of x in the earlier stated expressions to find the measure of the other angles:

x + 55 = 40 + 55 = 95

x + 65 = 40 + 65 = 105

x + 120 = 40 + 120 = 160

3 0
3 years ago
How would I answer this: How many minutes are in a 30-day month.... use vertical multiplication to get the right answer
Vikentia [17]

Answer:

43,200 minutes in 1months or 30 days

7 0
2 years ago
Read 2 more answers
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