Which is the best definition for the term "segment bisector"?

Mathematics

1 answer:

iren [92.7K]3 years ago

5 0

Answer:

a line that cuts a segment into two pieces of equal length.

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Christina had a total of $120 to spend while shopping. She already spent $68 on clothes . She wants to buy some video games that

avanturin [10]

Answer:

5 video games.

Step-by-step explanation:

In order to find the total number of video games that Tracy can buy with the money, she has we first need to find how much money she has left after buying the clothes. To do this we subtract the amount she spent on clothes by the initial total that Christina had.

120 - 68 = 52

Now that we know Christina has a remaining $52, we can divide this amount by the value of each individual game ($10) in order to calculate how many she can actually afford.

52 / 10 = 5.2

Finally, we can see that she can afford to only buy 5 video games.

4 0

3 years ago

Find all of the points on the x-axis which are 6 units from the point (−1, 1).

anyanavicka [17]

All the points that are 6 units from (-1, 1) are those on the circle
(x+1)^2 +(y-1)^2 = 36

For y=0, the two points of interest satisfy
(x+1)^2 +1 = 36
(x+1)^2 = 35 . . . . . . subtract 1
x+1 = ±√35
x = -1±√35

The points you seek are (-1-√35, 0) and (-1+√35, 0), about (-6.916, 0) and (4.916, 0).

5 0

3 years ago

Suppose that 75% of all trucks undergoing a brake inspection at a certain inspection facility pass the inspection. Consider grou

RSB [31]

Answer:

b) 0.2961

c) 0.2251

d) Mean = 11.25, Standard deviation = 1.667

Step-by-step explanation:

We are given the following information:

We treat trucks undergoing a brake inspection passin as a success.

P( trucks undergoing a brake inspection passes the test) = 75% = 0.75

a) Conditions for binomial probability distribution

There are n independent trial.
Each trial have a success probability p
The probability of success is same for all trials.

Then the number of trucks undergoing a brake inspection follows a binomial distribution, where

$P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}$

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 15

b) P(proportion of groups will between 8 and 10 trucks pass the inspection)

We have to evaluate:

$P(8\leq x \leq 10) = P(x = 8) + P(x = 9) + P(x = 10)\\= \binom{15}{8}(0.75)^8(1-0.75)^7 + \binom{15}{9}(0.75)^9(1-0.75)^6 + \binom{15}{10}(0.75)^{10}(1-0.75)^5\\= 0.0393 + 0.0917 + 0.1651\\= 0.2961$