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vovangra [49]
3 years ago
10

Assume that the weight of two year old babies have distribution that is approximately normal with a mean of 29 pounds and a stan

dard deviation of 3 pounds. what weight of two year old baby corresponds to 10th percentile?
Mathematics
1 answer:
Lana71 [14]3 years ago
5 0

Answer:

25.15 ponds is the weight of two year old baby corresponds to 10th percentile.      

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 29 pounds

Standard Deviation, σ = 3 pounds

We are given that the distribution of weight of two year old babies is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

We have to find the value of x such that the probability is 0.10

P(X < x)  

P( X < x) = P( z < \displaystyle\frac{x - 29}{3})=0.10  

Calculation the value from standard normal z table, we have,  

P(z < -1.282) = 0.10

\displaystyle\dfrac{x - 29}{3} = -1.282\\x = 25.154 \approx 25.15

Thus, 25.15 ponds is the weight of two year old baby corresponds to 10th percentile.

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Answer:

55 marbles.

Step-by-step explanation:

You multiply 3 and 6 and that equals 18. You multiply the numbers together because you know he has 3 bags of marbles and he has 6 marbles in each bag. Then you take the 18 marbles and add them together with the 37 marbles and you get 55 marbles.

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Step-by-step explanation:

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Click on the graph below to create a quadrilateral with vertices at the following points. (-4,9), (-8,9), (-9,7), (-9, 4)​
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...

To Find : plot the point and draw quadrilateral

Solution:

(-4,9)

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(-8,9)

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(-9,4)

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in her last computer game, Lucy scored 3×10⁷ points. The first time she tried the game, she scored 6×10³ points. How many times
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Step-by-step explanation:

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3 years ago
Prove that the triangle EDF is isosceles. Give reasons for your answer.
Gekata [30.6K]

Answer:

\triangle EDF is isosceles.

Step-by-step explanation:

Please have a look at the attached figure.

We are <u>given</u> the following things:

\angle EDF = y

\text{External }\angle DFG = 90 +\dfrac{y}{2}

Let us try to find out \angle E and \angle DFE. After that we will compare them.

<u>Finding </u>\angle DFE<u>:</u>

Side EG is a straight line so \angle GFE = 180

\angle GFE is sum of internal \angle DFE and external \angle DFG

\angle GFE = 180 = \angle DFE  + \angle DFG\\\Rightarrow 180 = \angle DFE + (90+\dfrac{y}{2})\\\Rightarrow \angle DFE = 180 - 90 - \dfrac{y}{2}\\\Rightarrow \angle DFE = 90 - \dfrac{y}{2} ....... (1)

<u>Finding </u>\angle E<u>:</u>

<u>Property of external angle:</u> External angle in a triangle is equal to the sum of two opposite internal angles of a triangle.

i.e. external \angle DFG = \angle E + \angle EDF

\Rightarrow 90+\dfrac{y}{2} = \angle E + y\\\Rightarrow \angle E = 90+\dfrac{y}{2}  -y\\\Rightarrow \angle E = 90-\dfrac{y}{2} ....... (2)

Comparing equations (1) and (2):

It can be clearly seen that:

\angle DFE = \angle E =90-\dfrac{y}{2}

The two angles of \triangle EDF are equal hence \triangle EDF is isosceles.

8 0
3 years ago
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