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3 years ago

Angle A is circumscribed about circle O. What is the measure of angle O? 46

Mathematics

1 answer:

masya89 [10]3 years ago

5 0

Answer:

m<O = 134°

Step-by-step explanation:

OC = OB = radius of the circle

AC = AB = tangents of circle O

m<C = m<B = 90°. (Tangent and a radius always form 90°)

m<A = 46°

Therefore,

m<O = 360° - (m<C + m<B + m<A) => sum of angles in a quadrilateral.

m<O = 360° - (90° + 90° + 46°)

m<O = 360° - 226°

m<O = 134°.

Measure of angle A = 134°

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5, 6, 10 Question: A. Determine whether the side lengths form a triangle. (explain your reasoning) B. If it is a triangle, deter

expeople1 [14]

Answer:

The right answer is C

Step-by-step explanation:

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3 years ago

Based on the model N(11541154,8686) describing steer weights, what are the cutoff values for a) the highest 10% of the we

yaroslaw [1]

Answer:

a) x = 1225.68

b) x = 1081.76

c) 1109.28 < x < 1198.72

Step-by-step explanation:

Given:

- Th random variable X for steer weight follows a normal distribution:

X~ N( 1154 , 86 )

Find:

a) the highest 10% of the weights?

b) the lowest 20% of the weights?

c) the middle 40% of the weights?

Solution:

We will compute the corresponding Z-value for highest cut off 10%:

Z @ 0.10 = 1.28

Z = (x-u) / sd

Where,

u: Mean of the distribution.

s.d: Standard deviation of the distribution.

1.28 = (x - 1154) / 86

x = 1.28*86 + 1154

x = 1225.68

We will compute the corresponding Z-value for lowest cut off 20%:

-Z @ 0.20 = -0.84

Z = (x-u) / sd

-0.84 = (x - 1154) / 86

x = -0.84*86 + 1154

x = 1081.76

We will compute the corresponding Z-value for middle cut off 40%:

Z @ 0.3 = -0.52

Z @ 0.7 = 0.52

[email protected] < x < [email protected]

-.52*86 + 1154 < x < 0.52*86 + 1154

1109.28 < x < 1198.72

7 0

3 years ago

It is known that the life of a particular auto transmission follows a normal distribution with mean 72,000 miles and standard de

scoray [572]

Answer:

a) $P(X$

$P(z$

b) $P(X>65000)=P(\frac{X-\mu}{\sigma}>\frac{65000-\mu}{\sigma})=P(Z>\frac{65000-72000}{12000})=P(z>-0.583)$

$P(z>-0.583)=1-P(Z$

c) $P(X>100000)=P(\frac{X-\mu}{\sigma}>\frac{100000-\mu}{\sigma})=P(Z>\frac{100000-72000}{12000})=P(z>2.33)$

$P(z>2.33)=1-P(Z$

Sicne this probability just represent 1% of the data we can consider this value as unusual.

d) $z=1.28$

And if we solve for a we got

$a=72000 +1.28*12000=87360$

So the value of height that separates the bottom 90% of data from the top 10% is 87360.

Step-by-step explanation:

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".

Part a

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

$X \sim N(72000,12000)$

Where $\mu=72000$ and $\sigma=12000$

We are interested on this probability

$P(X$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X$

And we can find this probability using excel or the normal standard table and we got:

$P(z$

Part b

$P(X>65000)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X>65000)=P(\frac{X-\mu}{\sigma}>\frac{65000-\mu}{\sigma})=P(Z>\frac{65000-72000}{12000})=P(z>-0.583)$

And we can find this probability using the complement rule and excel or the normal standard table and we got:

$P(z>-0.583)=1-P(Z$

Part c

$P(X>100000)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X>100000)=P(\frac{X-\mu}{\sigma}>\frac{100000-\mu}{\sigma})=P(Z>\frac{100000-72000}{12000})=P(z>2.33)$

And we can find this probability using the complement rule and excel or the normal standard table and we got:

$P(z>2.33)=1-P(Z$

Sicne this probability just represent 1% of the data we can consider this value as unusual.

Part d

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.1$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.9 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1

If we use condition (b) from previous we have this:

$P(X$

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=1.28$

And if we solve for a we got

$a=72000 +1.28*12000=87360$

So the value of height that separates the bottom 90% of data from the top 10% is 87360.

5 0

3 years ago

James is earning money to buy a $100 bicycle. For each dollar James earns, his mother has agreed to give him $1. So far, he has

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7 0

3 years ago

Awnser please please

astraxan [27]

They're all d).

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6 0

3 years ago