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MissTica
3 years ago
9

Which is not a common factor of 15, 45,and 90

Mathematics
1 answer:
Naddika [18.5K]3 years ago
7 0

Answer:

6/90/45/30/18/9/2

Step-by-step explanation:

15: 3/5/15/1

45: 45/1/9/5/15/3/

90: 90/1/9/10/45/2/15/6/3/30/5/18

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True or false? in order to inscribe a circle in a triangle, the circle's center must be placed at the circumcenter of the triang
Dvinal [7]

Answer:

<h2>False.</h2>

Step-by-step explanation:

A inscribe circle in a triangle means to draw the biggest circle possible inside such triangle. To do that perfectly, we first have to find the incenter of the circle, which is the intersection of all three internal bisector of the triangle, that point is the center of the inscribed circle.

Therefore, the statement is false.

In addition, a circumcenter allow to perfectly draw a circumscribed circle, which is outside the triangle, which is not the case here.

An image showing the inscribed circle is attached.

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3 years ago
Ajar contains 6 chocolate chip cookies and 9 peanut butter cookies. Richard grabs 3 cookies at random to pack in his lunch.
mel-nik [20]

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Step-by-step explanation:

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3 years ago
Janine has two choices for paying off her credit card balance. She can make payments of $186 per
Art [367]

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7 0
4 years ago
Read 2 more answers
What is the perimeter?
aev [14]

Answer:

38

Step-by-step explanation:

9²+10² = 38

5 0
3 years ago
The probabilities that a customer selects 1, 2, 3, 4, or 5 items at a store are 0.32, 0.12, 0.23, 0.18, and 0.15, respectively.
Andrew [12]

Answer:

a. This is a probability distribution as ∑pi=1

b. 2.72

c. 1.45

Step-by-step explanation:

a. To verify the probability distribution we simply add all given probabilities and see whether they sums up to 1 or not. According to definition of probability distribution sum of probabilities should be 1 so,

∑pi=0.32+0.12+0.23+0.18+0.15=1

Hence it is verified that given distribution is a probability distribution.

b. Mean number of items selected= E(x)= ∑x*p(x)

here x =1,2,3,4,5 and p(x)=0.32,0.12,0.23,0.18,0.15.

mean number of items=1*0.32+2*0.12+3*0.23+4*0.18+5*0.15=2.72

c. standard deviation of number of items=sqrt[(∑x²*p(x))-(∑x*p(x))²]

∑x²*p(x)=1*0.32+4*0.12+9*0.23+16*0.18+25*0.15=9.5

(∑x*p(x))²=(2.72)²=7.4

standard deviation of number of items=sqrt[(∑x²*p(x))-(∑x*p(x))²]=sqrt(9.5-7.4)=sqrt(2.1)=1.45

6 0
4 years ago
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