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3 years ago

HELP ME ASAP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics

1 answer:

Akimi4 [234]3 years ago

4 0

4. Answer 2 --> (-a, b) ; (a, b) ; (-c, d) ; (c, d)

5. Answer 1

6. Answer 4

7. Answer 2

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1. what value of x will make the equation below true 1/2(6x-10)+10=5x-13

soldi70 [24.7K]

Answer:

x = 9

Step-by-step explanation:

1/2(6x - 10) + 10 = 5x - 13

3x - 5 + 10 = 5x - 13

3x + 5 = 5x - 13

3x + 18 = 5x

18 = 2x

x = 9

3 0

3 years ago

Read 2 more answers

The team ran

juin [17]

The team ran 2 miles.

Step-by-step explanation:

Step 1; The track is $\frac{3}{4}$ miles long and the team ran 2 $\frac{2}{3}$ laps on it. To calculate the distance run we multiply the number of laps ran with the distance on each lap.

Distance run = Number of laps × Distance on each lap.

Step 2; We convert 2 $\frac{2}{3}$ from a mixed fraction into an improper fraction. To do this, we multiply the whole number with the denominator of the fraction and add with it the numerator of the same fraction whereas the denominator remains unchanged. To convert the fraction 2 $\frac{2}{3}$ = ((2 × 3) + 2) / 3 = $\frac{8}{3}$ .

Distance run = Number of laps × Distance on each lap = $\frac{8}{3}$ × $\frac{3}{4}$ = $\frac{8}{4}$ = 2 miles.

3 0

3 years ago

Read 2 more answers

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

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

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

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