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dalvyx [7]
3 years ago
8

Rozonda defines the diameter of a circle as a line segment passing through the center of the circle. Is Rozonda’s definition val

id?
Mathematics
2 answers:
larisa [96]3 years ago
6 0

Answer: Yes Rozonda's definition is valid

Step-by-step explanation:

denpristay [2]3 years ago
3 0

Answer:

Yes

Step-by-step explanation:

Rozanda's definition is valid because the actual definiton of a diameter within a circle any straight line segment that passes through the center of the circle and whose endpoints lie on the circle. It can also be defined as the longest chord of the circle. In more modern usage, the length of a diameter is also called the diameter.

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What is the value of x?<br><br> Enter your answer in the box.<br><br> x = in.
mr_godi [17]

Answer:

The answer is 24

I found this answer because the smaller triangles base is 5

and the bigger triangle is 15


5x3 = 15

8x3 = 24



7 0
3 years ago
Express f(x) in the form a(x-h)^2+k for f(x)=x^2-8x+21
Vlad [161]
X^2-8x+21

=x^2-2(4)(x) +16+5
=(x-4)^2+5

here a=1
h=4
k=5
3 0
3 years ago
Assume that a procedure yields a binomial distribution with a trial repeated n = 5 times. Use some form of technology to find th
Digiron [165]

Answer:

P(X = 0) = 0.0263

P(X = 1) = 0.1407

P(X = 2) = 0.3012

P(X = 3) = 0.3224

P(X = 4) = 0.1725

P(X = 5) = 0.0369

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

In this problem we have that:

n = 5, p = 0.517

Distribution

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{5,0}.(0.517)^{0}.(0.483)^{5} = 0.0263

P(X = 1) = C_{5,1}.(0.517)^{1}.(0.483)^{4} = 0.1407

P(X = 2) = C_{5,2}.(0.517)^{2}.(0.483)^{3} = 0.3012

P(X = 3) = C_{5,3}.(0.517)^{3}.(0.483)^{2} = 0.3224

P(X = 4) = C_{5,4}.(0.517)^{4}.(0.483)^{1} = 0.1725

P(X = 5) = C_{5,5}.(0.517)^{5}.(0.483)^{0} = 0.0369

6 0
4 years ago
The overhead reach distances of adult females are normally distributed with a mean of 205 cm and a standard deviation of 8.3 cm.
Gre4nikov [31]

a. Find the probability that an individual distance is greater than 214.30 cm

We find for the value of z score using the formula:

z = (x – u) / s

z = (214.30 – 205) / 8.3

z = 1.12

Since we are looking for x > 214.30 cm, we use the right tailed test to find for P at z = 1.12 from the tables:

P = 0.1314

 

b. Find the probability that the mean for 20 randomly selected distances is greater than 202.80 cm

We find for the value of z score using the formula:

z = (x – u) / s

z = (202.80 – 205) / 8.3

z = -0.265

Since we are looking for x > 202.80 cm, we use the right tailed test to find for P at z = -0.265 from the tables:

P = 0.6045

 

c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?

I believe this is because we are given the population standard deviation sigma rather than the sample standard deviation. So we can use the z test.

7 0
3 years ago
What order will result in a final output of 131,065 when the input is 64
Korolek [52]

Answer:

2080.69

Step-by-step explanation:

6 0
4 years ago
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