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Verdich [7]
2 years ago
11

The height of an adult male is known to be normally distributed with mean of 175 cm and standard deviation of 6 cm. The 20th per

centile (bottom 20%) of the distribution of heights is shorter than:
Mathematics
1 answer:
dusya [7]2 years ago
6 0

Answer:

The height below which 20% of the persons lie equals 169.95 cm

Step-by-step explanation:

Given that

Mean height = 175 cm

Standard deviation = 6 cm

Hence for the given question we need to find the standard variate factor ( Z factor) corresponding to the 20% of the area under the standard normal distribution curve

Using the standard normal distribution tables we have Z factor corresponding to an area of 20% equals ( Z = -0.84166)

Thus we have

Z=\frac{X-\bar{X }}{\sigma }\\\\\Rightarrow -0.84166=\frac{X-175}{6}\\\\\therefore X=-0.84166\times 6+175\\\\X=169.95cm

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–9x + 10y = –9 whats that in slope intercept form
kirill115 [55]
Slope intercept form is y = mx + b

so make the equation look like that. 

-9x + 10y = -9 ... add 9x on both sides 
10y = -9 + 9x 
10y = 9x - 9 
<span>
y = (9/10)x - (9/10)


I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
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4 0
2 years ago
Read 2 more answers
last night, julie's pet hamster zoe kept julie awake for at least an hour running on her exercise wheel and scratching at the co
borishaifa [10]

Answer: x+y\geq 1, x>\frac{1}{4} ,  x\geq 2y and y\geq 0

Step-by-step explanation:

Here,  x represents the number of hours Zoe spent running on her wheel and y represents the number of hours spent scratching her cage.

Julie was awoke for at least an hour running on her exercise wheel and scratching the of her cage.

⇒ x+y\geq 1

She ran on her wheel at least twice as long as she scratched at the corners of her cage.

⇒ x\geq 2y

Also, She spent more than 1/4 hour running on her wheel.

⇒ x>\frac{1}{4}

And, we know that number of hours can not be negative.

⇒  y\geq 0

Therefore, the complete system of inequality which shows the given situation is,

x+y\geq 1, x>\frac{1}{4} and x\geq 2y, y\geq 0

Note: the feasible region ( covered by the given system) is shown in the below graph.


6 0
2 years ago
Square $ABCD$ has area $200$. Point $E$ lies on side $\overline{BC}$. Points $F$ and $G$ are the midpoints of $\overline{AE}$ an
Charra [1.4K]

1. Consider square ABCD. You know that  

A_{ABCD}=AD^2=200,

then

AB=BC=CD=AD=10\sqrt{2}.

2. Consider traiangle AED. F is mipoint of AE and G is midpoint of DE, then FG is midline of triangle AED. This means that

FG=\dfrac{AD}{2}=\dfrac{10\sqrt{2} }{2}=5\sqrt{2}.

3. Consider trapezoid BFGC. Its area is

A_{BFGC}=\dfrac{FG+BC}{2}\cdot h, where h is the height of trapezoid and is equal to half of AB. Thus,

A_{BFGC}=\dfrac{FG+BC}{2}\cdot \dfrac{AB}{2}=\dfrac{5\sqrt{2}+10\sqrt{2}}{2}\cdot \dfrac{10\sqrt{2}}{2}=75.

4.

A_{BFGC}=A_{BFGE}+A_{EGC},\\A_{EGC}=A_{BFGC}-A_{BFGE}=75-34=41.

5. Note that angles EGC and CGD are supplementary and

\sin \angle CGD=\sin \angle EGC.

Then

A_{CGD}=\dfrac{1}{2}CG\cdot CD\cdot \sin \angle CGD=\dfrac{1}{2}CG\cdot EG\cdot \sin \angle CGE=A_{ACG}=41.

Answer: A_{CGD}=41.

8 0
2 years ago
A baseball player has 39 hits in 134 times at bat. how many hits must he get in the next 46 times at bat to finish the year with
faust18 [17]

Answer:

He must get 33 hits in his next 46 times at bat to finish the year with a .400 batting average

Step-by-step explanation:

The player has already batted 134 times and will still bat 46 times. So in the end of the year, he is going to have 134 + 46 = 180 at bats.

How many hits does he need to have to hit .400?

This is 40% of 180, which is 0.4*180 = 72.

He has already 39 hits, so in his next 46 at bats, he will need 72 - 39 = 33 hits.

4 0
2 years ago
Which function has a simplified base of 4RootIndex 3 StartRoot 4 EndRoot? f(x) = 2(RootIndex 3 StartRoot 16 EndRoot) Superscript
PIT_PIT [208]

Answer:

  f(x)=4\sqrt[3]{16}^{2x}

Step-by-step explanation:

We believe you're wanting to find a function with an equivalent base of ...

  4\sqrt[3]{4}\approx 6.3496

The functions you're looking at seem to be ...

  f(x)=2\sqrt[3]{16}^x\approx 2\cdot2.5198^x\\\\f(x)=2\sqrt[3]{64}^x=2\cdot 4^x\\\\f(x)=4\sqrt[3]{16}^{2x}\approx 4\cdot 6.3496^x\ \leftarrow\text{ this one}\\\\f(x)=4\sqrt[3]{64}^{2x}=4\cdot 16^x

The third choice seems to be the one you're looking for.

9 0
2 years ago
Read 2 more answers
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