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spayn [35]
3 years ago
15

Find the given measure

Mathematics
1 answer:
Vanyuwa [196]3 years ago
7 0

Answer:

80°

Step-by-step explanation:

The angle on the circle ∠DBC is half the measure of the arc it intercepts, thus

m CD = 2 × 40° = 80°

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in 1950, the population of a city was about 800,000 people. Over the next 50 year, the population increased to 1.3 million.Find
jasenka [17]

Answer:

62.5% increase

Step-by-step explanation:

First step : find the difference between two numbers, in this case, it's 1300000 - 800000 = 500000.

Second Step: Take the difference, 500000, and divide by the original number: 500000/800000  = 0.625.

Last, multiply the number above by 100: 0.625*100 = 62.5%. You're done! You calculated difference of a number in percent, and the answer is a percentage increase of 62.5%.

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3 years ago
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Which of the following reasons can be used to justify statement #3 in the proof?
mixer [17]

c because it is accuet.

hope it helps

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3 years ago
Regina writes the expression y + 9 · ¼. Which expression is equivalent to the one Regina writes?
Alika [10]

Answer:

Regina writes the expression y +9.3/4

Using The commutative property

a + b = b + a

= 9.3/4  + y

= (9.3/4) + y

Step-by-step explanation:

Regina writes the expression y +9.3/4

Using The commutative property

a + b = b + a

= 9.1/4  + y

= (9.1/4) + y

5 0
2 years ago
I would much appreciate the explanation for number 20 cuz ion’ get it
cestrela7 [59]

Answer:

Which is number 20?

Step-by-step explanation:

Which is number 20?

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3 years ago
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Question Help Suppose that the lifetimes of light bulbs are approximately normally​ distributed, with a mean of 5656 hours and a
koban [17]

Answer:

a)3.438% of the light bulbs will last more than 6262 hours.

b)11.31% of the light bulbs will last 5252 hours or less.

c) 23.655% of the light bulbs are going to last between 5858 and 6262 hours.

d) 0.12% of the light bulbs will last 4646 hours or less.

Step-by-step explanation:

Normally distributed problems can be solved by the z-score formula:

On a normaly distributed set with mean \mu and standard deviation \sigma, the z-score of a value X is given by:

Z = \frac{X - \mu}{\sigma}

After we find the value of Z, we look into the z-score table and find the equivalent p-value of this score. This is the probability that a score will be LOWER than the value of X.

In this problem, we have that:

The lifetimes of light bulbs are approximately normally​ distributed, with a mean of 5656 hours and a standard deviation of 333.3 hours.

So \mu = 5656, \sigma = 333.3

(a) What proportion of light bulbs will last more than 6262 ​hours?

The pvalue of the z-score of X = 6262 is the proportion of light bulbs that will last less than 6262. Subtracting 100% by this value, we find the proportion of light bulbs that will last more than 6262 hours.

Z = \frac{X - \mu}{\sigma}

Z = \frac{6262 - 5656}{333.3}

Z = 1.82

Z = 1.81 has a pvalue of .96562. This means that 96.562% of the light bulbs are going to last less than 6262 hours. So

P = 100% - 96.562% = 3.438% of the light bulbs will last more than 6262 hours.

​(b) What proportion of light bulbs will last 5252 hours or​ less?

This is the pvalue of the zscore of X = 5252

Z = \frac{X - \mu}{\sigma}

Z = \frac{5252- 5656}{333.3}

Z = -1.21

Z = -1.21 has a pvalue of .1131. This means that 11.31% of the light bulbs will last 5252 hours or less.

(c) What proportion of light bulbs will last between 5858 and 6262 ​hours?

This is the pvalue of the zscore of X = 6262 subtracted by the pvalue of the zscore X = 5858

For X = 6262, we have that Z = 1.81 with a pvalue of .96562.

For X = 5858

Z = \frac{X - \mu}{\sigma}

Z = \frac{5858- 5656}{333.3}

Z = 0.61

Z = 0.61 has a pvalue of .72907.

So, the proportion of light bulbs that will last between 5858 and 6262 hours is

P = .96562 - .72907 = .23655

23.655% of the light bulbs are going to last between 5858 and 6262 hours.

​(d) What is the probability that a randomly selected light bulb lasts less than 4646 ​hours?

This is the pvalue of the zscore of X = 4646

Z = \frac{X - \mu}{\sigma}

Z = \frac{4646- 5656}{333.3}

Z = -3.03

Z = -3.03 has a pvalue of .0012. This means that 0.12% of the light bulbs will last 4646 hours or less.

5 0
3 years ago
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