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NARA [144]
2 years ago
8

In ΔABC, AD and BE are the angle bisectors of ∠A and ∠B and DE ║ AB . If m∠ADE is with 34° smaller than m∠CAB, find the measures

of the angles of ΔADE.

Mathematics
1 answer:
Charra [1.4K]2 years ago
7 0

Answer:

34°

Step-by-step explanation:

If m∠ADE is with 34° smaller than m∠CAB, then denote

m∠ADE=x°,

m∠CAB=(x+34)°.

Since  DE ║ AB, then

m∠ADE=m∠DAB=x°.

AD is angle A bisector, then

m∠EAD=m∠DAB=x°.

Thus,

m∠CAB=m∠CAD+m∠DAB=(x+x)°=2x°.

On the other hand,

m∠CAB=(x+34)°,

then

2x°=(x+34)°,

m∠ADE=x°=34°.

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y(.82)*1.08

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You multiply by .82 to get the cost of the game after the discount and then by 1.08 to add the tax and total amount.  

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Add whole numbers: 2 + 6 + 8 = 16
Add fractions:

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a golfer needs to hit a ball a distance of 500 feet, but there is a 60-foot tall tree that is 100 feet in front of the point whe
jok3333 [9.3K]

We can create a parabola equation of the trajectory using the vertex form:

y = a (x – h)^2 + k

 

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k = 120

 

Therefore:

 y = a (x – 250)^2 + 120

 

At the initial point, x = 0, y = 0, so we can solve for a:

0 = a (0 – 250)^2 + 120

0 = a (62,500) + 120

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So the whole equation is:

y = -0.00192 (x – 250)^2 + 120

 

So find for y when the golf ball is above the tree, x = 400:

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2 years ago
A random sample of 36 observations has been drawn from a normal distribution with mean 50 and standard deviation 12. Find the pr
Gre4nikov [31]

Answer:

z= \frac{47 -50}{\frac{12}{\sqrt{36}}}=-1.5

z= \frac{53 -50}{\frac{12}{\sqrt{36}}}=1.5

And using a calculator, excel ir the normal standard table we have that:

P(47 \leq \bar X \leq 53) =P(-1.5 \leq Z \leq 1.5)

And we can calculate the probability like this:P(-1.5 \leq Z \leq 1.5) = P(zStep-by-step explanation:

A random sample of 36 observations has been drawn from a normal distribution with mean 50 and standard deviation 12. Find the probability that the sample mean is in the interval 47<=X<53. Is the assumption of normality important. Why?

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

X \sim N(50,12)  

Where \mu=50 and \sigma=12

Since the distribution for X is normal then we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

We can find the probability required like this:

z= \frac{47 -50}{\frac{12}{\sqrt{36}}}=-1.5

z= \frac{53 -50}{\frac{12}{\sqrt{36}}}=1.5

And using a calculator, excel ir the normal standard table we have that:

P(47 \leq \bar X \leq 53) =P(-1.5 \leq Z \leq 1.5)

And we can calculate the probability like this:

P(-1.5 \leq Z \leq 1.5) = P(z

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