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krok68 [10]
3 years ago
10

5. Which step is included in the construction of inscribed polygons? (5 points)

Mathematics
1 answer:
RideAnS [48]3 years ago
5 0

Answer:

Your answer is D

Step-by-step explanation:

Your welcome

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Three poultry farms B1, B2, and B3 supply chicken eggs to the same local market. The market performs some quality testing on the
ipn [44]

Answer:

a. 0.58

b. 0.78

Step-by-step explanation:

a. The probability of egg come from B1 or B2

P(B1) = 3000/10000 = 0.3

P(B2) = 4000/10000 = 0.4

P(P1 ∪ B2) = 0.3 + 0.4 -(0.3)(0.4)

P(P1 ∪ B2) = 0.7 - 0.12

P(P1 ∪ B2) = 0.58

b. The probability that the market received an egg that is acceptable

P(received an egg that is acceptable) = P(B1 acceptable) + P(B2 acceptable) + P(B3 acceptable)

P(received an egg that is acceptable) = 0.80*3000 + 0.90*4000 + 0.60*3000 / 10000

P(received an egg that is acceptable) = 2400 + 3600 + 1800 / 10000

P(received an egg that is acceptable) = 7800 / 10000

P(received an egg that is acceptable) = 0.78

4 0
3 years ago
Graph x^4-3x^2+x. Identify the x-intercepts and the points where the local maximums and local minimums occur. Determine the inte
Brums [2.3K]

Functions, equations and tables can be represented using graphs

  • The x-intercepts are: 0, 0.35 and 1.53
  • The local maximum is at point (0.17, 0.08), and the local minimum is at point (1.13, -1.07)

The expression is given as:

x^4-3x^2+x

Rewrite the expression as a function

f(x) =  x^4-3x^2+x

See attachment for the graph of the function.

<h3>The x-intercepts</h3>

This is the point where the graph cross the x-axis.

From the attached graph, the x-intercepts are: 0, 0.35 and 1.53

<h3>The local maximum</h3>

From the attached graph, the local maximum is at point (0.17, 0.08)

<h3>The local minimum</h3>

From the attached graph, the local minimum is at point (1.13, -1.07)

<h3>Intervals</h3>
  • The graph increases at interval (-1.30, 0) and (1,13, infinity)
  • The graph decreases at interval (- infinity, 1.30) and (0.35, 1.13)

Read more about graphs and functions at:

brainly.com/question/13136492

4 0
2 years ago
A 6 sided die numbered 1 to 6 is rolled . Find the probability that in the roll of the results is an odd number or less than 4
stellarik [79]

Answer:

4/6 or 66%

Step-by-step explanation:

Odd numbers from 1 to  6 are: 1, 3, 5

Numbers less than 4 are: 1, 2, 3

So there are  4/ 6 possible outcomes.

So the probability is 4/6 or 66%.

3 0
3 years ago
Suppose a geyser has a mean time between eruptions of 72 minutes. Let the interval of time between the eruptions be normally dis
nikitadnepr [17]

Answer:

(a) The probability that a randomly selected time interval between eruptions is longer than 82 ​minutes is 0.3336.

(b) The probability that a random sample of 13-time intervals between eruptions has a mean longer than 82 ​minutes is 0.0582.

(c) The probability that a random sample of 34 time intervals between eruptions has a mean longer than 82 ​minutes is 0.0055.

(d) Due to an increase in the sample size, the probability that the sample mean of the time between eruptions is greater than 82 minutes decreases because the variability in the sample mean decreases as the sample size increases.

(e) The population mean must be more than 72​, since the probability is so low.

Step-by-step explanation:

We are given that a geyser has a mean time between eruptions of 72 minutes.

Also, the interval of time between the eruptions be normally distributed with a standard deviation of 23 minutes.

(a) Let X = <u><em>the interval of time between the eruptions</em></u>

So, X ~ N(\mu=72, \sigma^{2} =23^{2})

The z-score probability distribution for the normal distribution is given by;

                            Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean time = 72 minutes

           \sigma = standard deviation = 23 minutes

Now, the probability that a randomly selected time interval between eruptions is longer than 82 ​minutes is given by = P(X > 82 min)

       P(X > 82 min) = P( \frac{X-\mu}{\sigma} > \frac{82-72}{23} ) = P(Z > 0.43) = 1 - P(Z \leq 0.43)

                                                           = 1 - 0.6664 = <u>0.3336</u>

The above probability is calculated by looking at the value of x = 0.43 in the z table which has an area of 0.6664.

(b) Let \bar X = <u><em>sample mean time between the eruptions</em></u>

The z-score probability distribution for the sample mean is given by;

                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean time = 72 minutes

           \sigma = standard deviation = 23 minutes

           n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between eruptions has a mean longer than 82 ​minutes is given by = P(\bar X > 82 min)

       P(\bar X > 82 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{82-72}{\frac{23}{\sqrt{13} } } ) = P(Z > 1.57) = 1 - P(Z \leq 1.57)

                                                           = 1 - 0.9418 = <u>0.0582</u>

The above probability is calculated by looking at the value of x = 1.57 in the z table which has an area of 0.9418.

(c) Let \bar X = <u><em>sample mean time between the eruptions</em></u>

The z-score probability distribution for the sample mean is given by;

                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean time = 72 minutes

           \sigma = standard deviation = 23 minutes

           n = sample of time intervals = 34

Now, the probability that a random sample of 34 time intervals between eruptions has a mean longer than 82 ​minutes is given by = P(\bar X > 82 min)

       P(\bar X > 82 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{82-72}{\frac{23}{\sqrt{34} } } ) = P(Z > 2.54) = 1 - P(Z \leq 2.54)

                                                           = 1 - 0.9945 = <u>0.0055</u>

The above probability is calculated by looking at the value of x = 2.54 in the z table which has an area of 0.9945.

(d) Due to an increase in the sample size, the probability that the sample mean of the time between eruptions is greater than 82 minutes decreases because the variability in the sample mean decreases as the sample size increases.

(e) If a random sample of 34-time intervals between eruptions has a mean longer than 82 ​minutes, then we conclude that the population mean must be more than 72​, since the probability is so low.

6 0
3 years ago
Triangle JJJ is shown below. James drew a scaled version of Triangle JJJ using a scale factor of 444 and labeled it Triangle KKK
Blababa [14]
How am I so supposed to solve the question without the picture?
3 0
3 years ago
Read 2 more answers
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