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Amanda [17]
3 years ago
11

What expression is equivalent to 5x-2

Mathematics
1 answer:
NikAS [45]3 years ago
4 0

Answer:

x+3

Step-by-step explanation:

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this is step three of mathematical induction can somebody help me solve the rest, you only work on the left side to make it equa
kvv77 [185]

Answer:

Step 1(Base step) − It proves that a statement is true for the initial value. Step 2(Inductive step) − It proves that if the statement is true for the nth iteration (or number n), then it is also true for (n+1)th iteration ( or number n+1).

Step-by-step explanation:

3 0
3 years ago
Four buses carrying 146 high school students arrive to Montreal. The buses carry, respectively, 32, 44, 28, and 42 students. One
Naily [24]

Answer:

The expected value of X is E(X)=\frac{2754}{73} \approx 37.73 and the variance of X is Var(X)=\frac{226192}{5329} \approx 42.45

The expected value of Y is E(Y)=\frac{73}{2} \approx 36.5 and the  variance of Y is Var(Y)=\frac{179}{4} \approx 44.75

Step-by-step explanation:

(a) Let X be a discrete random variable with set of possible values D and  probability mass function p(x). The expected value, denoted by E(X) or \mu_x, is

E(X)=\sum_{x\in D} x\cdot p(x)

The probability mass function p_{X}(x) of X is given by

p_{X}(28)=\frac{28}{146} \\\\p_{X}(32)=\frac{32}{146} \\\\p_{X}(42)=\frac{42}{146} \\\\p_{X}(44)=\frac{44}{146}

Since the bus driver is equally likely to drive any of the 4 buses, the probability mass function p_{Y}(x) of Y is given by

p_{Y}(28)=p_{Y}(32)=p_{Y}(42)=p_{Y}(44)=\frac{1}{4}

The expected value of X is

E(X)=\sum_{x\in [28,32,42,44]} x\cdot p_{X}(x)

E(X)=28\cdot \frac{28}{146}+32\cdot \frac{32}{146} +42\cdot \frac{42}{146} +44 \cdot \frac{44}{146}\\\\E(X)=\frac{392}{73}+\frac{512}{73}+\frac{882}{73}+\frac{968}{73}\\\\E(X)=\frac{2754}{73} \approx 37.73

The expected value of Y is

E(Y)=\sum_{x\in [28,32,42,44]} x\cdot p_{Y}(x)

E(Y)=28\cdot \frac{1}{4}+32\cdot \frac{1}{4} +42\cdot \frac{1}{4} +44 \cdot \frac{1}{4}\\\\E(Y)=146\cdot \frac{1}{4}\\\\E(Y)=\frac{73}{2} \approx 36.5

(b) Let X have probability mass function p(x) and expected value E(X). Then the variance of X, denoted by V(X), is

V(X)=\sum_{x\in D} (x-\mu)^2\cdot p(x)=E(X^2)-[E(X)]^2

The variance of X is

E(X^2)=\sum_{x\in [28,32,42,44]} x^2\cdot p_{X}(x)

E(X^2)=28^2\cdot \frac{28}{146}+32^2\cdot \frac{32}{146} +42^2\cdot \frac{42}{146} +44^2 \cdot \frac{44}{146}\\\\E(X^2)=\frac{10976}{73}+\frac{16384}{73}+\frac{37044}{73}+\frac{42592}{73}\\\\E(X^2)=\frac{106996}{73}

Var(X)=E(X^2)-(E(X))^2\\\\Var(X)=\frac{106996}{73}-(\frac{2754}{73})^2\\\\Var(X)=\frac{106996}{73}-\frac{7584516}{5329}\\\\Var(X)=\frac{7810708}{5329}-\frac{7584516}{5329}\\\\Var(X)=\frac{226192}{5329} \approx 42.45

The variance of Y is

E(Y^2)=\sum_{x\in [28,32,42,44]} x^2\cdot p_{Y}(x)

E(Y^2)=28^2\cdot \frac{1}{4}+32^2\cdot \frac{1}{4} +42^2\cdot \frac{1}{4} +44^2 \cdot \frac{1}{4}\\\\E(Y^2)=196+256+441+484\\\\E(Y^2)=1377

Var(Y)=E(Y^2)-(E(Y))^2\\\\Var(Y)=1377-(\frac{73}{2})^2\\\\Var(Y)=1377-\frac{5329}{4}\\\\Var(Y)=\frac{179}{4} \approx 44.75

8 0
3 years ago
Can someone please answer this question fast ?<br> the question is in the picture down
Lena [83]

Answer:

Yes, Always

Step-by-step explanation:

If M is the midpoint of line KL, then M, K, and L are collinear.

As you notice in the diagram, to be collinear means part of the same line.

please mark brainliest

4 0
2 years ago
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3) Which point is located at (6, 3)?<br><br>A <br>B <br>C <br>D​
maks197457 [2]

Answer:

C

Step-by-step explanation:

because its at 6,3

8 0
2 years ago
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What is 1000 to the power of 4
crimeas [40]
1*10^12
Which means like 11 0's at the end of the 10
8 0
3 years ago
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