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Olegator [25]
3 years ago
14

The area under a particular normal curve between 6 and 8 is 0.695. A normally distributed variable has the same mean and standar

d deviation as the parameters for this normal curve. What percentage of all possible observations of the variable lie between 6 and 8​?
Mathematics
1 answer:
TiliK225 [7]3 years ago
4 0

Answer:

69.5%

Step-by-step explanation:

A feature of the normal distribution is that this is completely determined by its mean and standard deviation, therefore, if two normal curves have the same mean and standard deviation we can be sure that they are the same normal curve. Then, the probability of getting a value of the normally distributed variable between 6 and 8 is 0.695. In practice we can say that if we get a large sample of observations of the variable, then, the percentage of all possible observations of the variable that lie between 6 and 8 is 100(0.695)% = 69.5%.

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I’ll mark brainliest if correct lol
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Answer:

10 seconds

Step-by-step explanation:

mark two points on the line the use the slope rule to calculate it you'll find it's -5 m/s

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3 years ago
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After prime planting season was over, a horticulturist sold lilac bushes for $15.00. If the original price was $39.00, what is t
hoa [83]
Find what percent of $39.00 the sale price was.
=> 15/39*100=38.4%
now subtract from 100% to get the mark down percentage.
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Final answer:
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4 0
3 years ago
Data consistent with summary quantities in the article "Effects of Fast-Food Consumption on Energy Intake and Diet Quality Among
Olegator [25]

Answer:

a) X[bar]₁=  1839.20 cal

b) X[bar]₂= 1779.07 cal

c) S₁= 386.35 cal

Step-by-step explanation:

Hello!

You have two independent samples,

Sample 1: n₁= 15 children that did not eat fast food.

Sample 2: n₂= 15 children that ate fast food.

The study variables are:

X₁: Calorie consumption of a kid that does not eat fast food in one day.

X₂: Calorie consumprion of a kid that eats fast food in one day.

a)

The point estimate of the population mean is the sample mean

X[bar]₁= (∑X₁/n₁) = (27588/15)= 1839.20 cal

b)

X[bar]₂= (∑X₂/n₂)= (26686/15)= 1779.07 cal

c)

To calculate the sample standard deiation, you have to calculate the sample variance first:

S₁²= \frac{1}{n_1-1}[∑X₁² - (( ∑X₁)²/n₁)]= \frac{1}{14} + [52829538 - (\frac{27588^{2} }{15}) = 149263.4571 cal²

S₁= 386.35 cal

I hope it helps!

7 0
3 years ago
I need help with this one too
natali 33 [55]

Hey there! :)

Answer:

y = 11.

Step-by-step explanation:

Set both of the equations equal:

5x - 9 = x² - 3x + 7

Rearrange the equation:

0 = x² - 3x - 5x  + 9  + 7

Combine like terms:

0 = x² - 8x + 16

Factor:

0 = (x -4)²

Solve for x:

0 = x - 4

x = 4.

Plug this into an equation to solve for 'y':

y = 5(4) - 9

y = 20 - 9

y = 11.

3 0
3 years ago
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Answer:B) 150

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