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MaRussiya [10]
3 years ago
10

Which is a true statement about any two chords that are the same distance from the center of a circle?

Mathematics
2 answers:
stepladder [879]3 years ago
8 0
If two cords are the same distance from the center of a circle than both cords are congruent 
The rEaL Qb9 ironic amr
2 years ago
Apex- They are congruent
goldenfox [79]3 years ago
3 0

Answer C. If two chords are the exact same distance from the center of the circle then those two chords must be congruent.

The rEaL Qb9 ironic amr
2 years ago
Apex- They are congruent
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Why is this correct?
Vanyuwa [196]

Answer: A= independent variable W = dependent variable  

Step-by-step explanation:

w= a/25

7 0
3 years ago
20 Point!
ludmilkaskok [199]

determinant: \sqrt{b^2-4ac}

(a) x^2+4x+5=0\\D=4^2-4\cdot1\cdot5={-4}\\D

D<0 means there are no real roots. there are two complex roots with imaginary components.

(b) D=16+20=36>0

D>0 means there are two real roots

(c) D = 20^2-4*4*25 = 0

D=0 means there is one real root with multiplicity 2


7 0
3 years ago
Help meeeeeeeeeee please
daser333 [38]

Answer:

6

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
An instructor who taught two sections of engineering statistics last term, the first with 25 students and the second with 35, de
Amiraneli [1.4K]

Answer:

a) P=0.1721

b) P=0.3528

c) P=0.3981

Step-by-step explanation:

This sampling can be modeled by a binominal distribution where p is the probability of a project to belong to the first section and q the probability of belonging to the second section.

a) In this case we have a sample size of n=15.

The value of p is p=25/(25+35)=0.4167 and q=1-0.4167=0.5833.

The probability of having exactly 10 projects for the second section is equal to having exactly 5 projects of the first section.

This probability can be calculated as:

P=\frac{n!}{(n-k)!k!}p^kq^{n-k}= \frac{15!}{(10)!5!}\cdot 0.4167^5\cdot0.5833^{10}=0.1721

b) To have at least 10 projects from the 2nd section, means we have at most 5 projects for the first section. In this case, we have to calculate the probability for k=0 (every project belongs to the 2nd section), k=1, k=2, k=3, k=4 and k=5.

We apply the same formula but as a sum:

P(k\leq5)=\sum_{k=0}^{5}\frac{n!}{(n-k)!k!}p^kq^{n-k}

Then we have:

P(k=0)=0.0003\\P(k=1)=0.0033\\P(k=2)=0.0165\\P(k=3)=0.0511\\P(k=4)=0.1095\\P(k=5)=0.1721\\\\P(k\leq5)=0.0003+0.0033+0.0165+0.0511+0.1095+0.1721=0.3528

c) In this case, we have the sum of the probability that k is equal or less than 5, and the probability tha k is 10 or more (10 or more projects belonging to the 1st section).

The first (k less or equal to 5) is already calculated.

We have to calculate for k equal to 10 or more.

P(k\geq10)=\sum_{k=10}^{15}\frac{n!}{(n-k)!k!}p^kq^{n-k}

Then we have

P(k=10)=0.0320\\P(k=11)=0.0104\\P(k=12)=0.0025\\P(k=13)=0.0004\\P(k=14)=0.0000\\P(k=15)=0.0000\\\\P(k\geq10)=0.032+0.0104+0.0025+0.0004+0+0=0.0453

The sum of the probabilities is

P(k\leq5)+P(k\geq10)=0.3528+0.0453=0.3981

8 0
3 years ago
What is the factored form of the expression?
Georgia [21]

Answer: The answer is (d) (4x+7)(7x+8).


Step-by-step explanation: Given equation to be factored is

28x^2+81x+56.

The given expression is quadratic, so we can factor it into two linear factors, by using the following factorisation rule.

x^2+(a+b)x+ab=(x+a)(x+b).

So, the given equation can be factorised using the above rule. The factorisation is as follows.

28x^2+81x+56\\\\=28x^2+49x+32x+56\\\\=7x(4x+7)+8(4x+7)\\\\=(4x+7)(7x+8).

Thus, the correct option is (d).





7 0
3 years ago
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