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2 years ago

What is the difference of the means of the distributions?

Mathematics

1 answer:

alexdok [17]2 years ago

4 0

The different of the means of the distribution will be 15. Then the correct option is A.

<h3>What is a normal distribution?</h3>

The Gaussian Distribution is another name for it. The most significant continuous probability distribution is this one. Because the curve resembles a bell, it is also known as a bell curve.

the graph is shown below.

Then the different of the means of the distribution will be

⇒ 45 - 30

⇒ 15

Then the correct option is A.

More about the normal distribution link is given below.

brainly.com/question/12421652

#SPJ1

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Terry Sheehan owns property assessed at $390,000 and pays a property tax rate of 1.69 per hundred dollars of assessed value. He

34kurt

Answer:

$1689

Step-by-step explanation:

Excess of rent amount over annual property tax bill = annual rent - annual property tax

annual property tax = [(assessed property value / 100) x 1.69]

( $390,000 / 100) x 1.69 = $6591

Annual rent = $690 x 12 = $8280

$8280 - $6591 = $1689

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3 years ago

Linda ran 2 mile in 1/3 hour. How many <br> miles did she run in one hour?

Semenov [28]

She would’ve ran 6 miles in an hour.

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3 years ago

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How many bowls of cereal did he eat pls and thx

miv72 [106K]

36 bowls - since with one gallon he can eat 12 bowls (as demonstrated by 1/12), you multiply 12 by 3 to get 36 bowls.

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4 years ago

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Jane wishes to bake an apple pie for dessert. The baking instructions say that she should bake the pie in an oven at a constant

pychu [463]

Answer:

A=152

K= -Ln(0.5)/14

Step-by-step explanation:

You can obtain two equations with the given information:

T(14 minutes) = 114◦C

T(28 minutes)=152◦C

Therefore, you have to replace t=14, T=114 and t=28, T=152 in the given equation:

$114=190-Ae^{-14k} (I) \\152=190-Ae^{-28k}(II)$

Applying the following property of exponentials numbers in (II):

$e^{a}.e^{b}=e^{a+b}$

Therefore $e^{-28k}$ can be written as $e^{-14k}.e^{-14k}$

$152=190-Ae^{-14k}.e^{14k}$

Replacing (I) in the previous equation:

$152=190-76e^{-14k}$

Solving for k:

Subtracting 190 both sides, dividing by -76:

$0.5=e^{-14k}$

Applying the base e logarithm both sides:

Ln(0.5)= -14k

Dividing by -14:

k= -Ln(0.5)/14

Replacing k in (I) and solving for A:

$Ae^{-14(-Ln(0.5)/14)}=76\\Ae^{Ln(0.5)} =76\\A(0.5)=76$

Dividing by 0.5

A=152

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3 years ago

Given: -16 = m/5 -18; <br> Prove: m=10

spayn [35]

Answer:

I have proven m is 10. have a good day and give brainliest if possible.

3 0

2 years ago