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JulsSmile [24]
3 years ago
10

Use substitution to solve the system of equations. x-2y=1 3x-6y=3

Mathematics
1 answer:
Ivahew [28]3 years ago
7 0
Change first equation
x - 2y = 1
add 2y to both side
x = 2y + 1
Plug into second equation
3(2y+1) - 6y = 3
Distributive property
6y + 3 - 6y = 3
y = 0, plug into first equation
x - 2(0) = 1, x = 1
Solution: x = 1, y = 0
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Choose the kind(s) of symmetry: point, line, plane, or none. (B)
max2010maxim [7]

Answer:

Hello there. The guy who answered before was only half correct. the correct answer is Line and Plane. Also just did this got it correct.

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3 years ago
I have to reflect a triangle over the line of reflection n: y = 0. What does that mean?
Natasha_Volkova [10]

Answer:

basically just reflect it over the y-axis since it equals 0

6 0
3 years ago
Find the mean,median,mode and range for each set of data its 23,27,24,26,26,24,26,24
lilavasa [31]

Mean is the average of the data set, which is found by adding all values together and then dividing that sum by the number of data values.

Median is the middle number of the data set, and can be found by ordering the values from least to greatest. If there are two middle numbers, the average of the two would be the median.

Mode is the number that shows up most frequently in the data set.

Range is found by subtracting the lowest number from the highest number in the data set.

<u>Set 1</u>

Mean:

18+20+22+11+19+18+18=126

126÷7=18

Median:

11, 18, 18, 18, 19, 20, 22 →  18

Mode: 18

Range: 22-11=11

<u>Set 2</u>

Mean: 23+27+24+26+26+24+26+24=200

200÷8=25

Median:

23, 24, 24, 24, 26, 26, 26, 27 → 24+26=50 → 50÷2=25

Mode: 24 and 26

Range: 27-23=4

4 0
3 years ago
Read 2 more answers
the name Joe is very common at a school in one out of every ten students go by the name. If there are 15 students in one class,
kumpel [21]

Using the binomial distribution, it is found that there is a 0.7941 = 79.41% probability that at least one of them is named Joe.

For each student, there are only two possible outcomes, either they are named Joe, or they are not. The probability of a student being named Joe is independent of any other student, hence, the <em>binomial distribution</em> is used to solve this question.

<h3>Binomial probability distribution </h3>

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

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

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

In this problem:

  • One in ten students are named Joe, hence p = \frac{1}{10} = 0.1.
  • There are 15 students in the class, hence n = 15.

The probability that at least one of them is named Joe is:

P(X \geq 1) = 1 - P(X = 0)

In which:

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

P(X = 0) = C_{15,0}.(0.1)^{0}.(0.9)^{15} = 0.2059

Then:

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.2059 = 0.7941

0.7941 = 79.41% probability that at least one of them is named Joe.

To learn more about the binomial distribution, you can take a look at brainly.com/question/24863377

8 0
2 years ago
Which shows all the names that apply to the figure?
Sloan [31]
B. parallelogram , rhombus , and rectangle
6 0
3 years ago
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