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3 years ago

Click and drag like terms onto each other to simplify fully. 1+x^2-3+2y+7x^2+3+3y

Mathematics

2 answers:

Rashid [163]3 years ago

6 0

Answer:

8x^2 + 5y + 1

Step-by-step explanation:

1 + x^2 - 3 + 2y + 7x^2 + 3 + 3y

group like terms

= x^2 + 7x^2 + 2y + 3y + 1 - 3 + 3

add similar items

= 8x^2 + 2y + 3y + 1 - 3 + 3

add similar items

= 8x^2 + 5y + 1 - 3 + 3

= 8x^2 + 5y + 1

Sidana [21]3 years ago

6 0

Answer:

$\Huge \boxed{8x^2 +5y+1}$

$\rule[225]{225}{2}$

Step-by-step explanation:

$1+x^2-3+2y+7x^2+3+3y$

Grouping like terms.

$7x^2 +x^2+2y +3y +3+1 -3$

$(7x^2 +x^2)+(2y +3y) +(3+1 -3)$

Combining like terms.

$8x^2 +5y+1$

$\rule[225]{225}{2}$

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Answer:

-6 < x ≤ 6

Step-by-step explanation:

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3 years ago

Read 2 more answers

Help me with this I don't get it help4help?

Aleks [24]

1) Divide 3/4 by 1/2. Remember how to divide fractions. Switch the problem into a multiplication problem.
3/4 x 2/1 = 3/2 (so you could only have 1 and 1/2 servings with 1/2 pound of chicken in each serving. 

2) Set up a little algebra problem. Let x = the multiplier of the original 3/4 pounds of chicken needed to make two full servings with 1/2 pound of chicken in each. 
(3/4)x = 2(1/2) (Found this online V(0-0)v)

6 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

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

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

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