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2 years ago

Is 45+12.5x greater than less than or equal to 75+7.5x

Mathematics

1 answer:

tiny-mole [99]2 years ago

7 0

Answer:

$45+12.5x < 75+7.5x$

Step-by-step explanation:

set each problem to 0 and solve.

Take the answer and plug it in for x. the 1st problem is less than the 2nd.

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When truckers are on long-haul drives, their driving logs must reflect their average speed. Average speed is the total distance

bekas [8.4K]

a) $v=\frac{d_{tot}}{t_{tot}}=\frac{(3 h)(60 mph)+20 mi}{3 h +t_2}$

The average speed is equal to the ratio between the total distance ( $d_{tot}$ and the total time taken ( $t_{tot}$ ):

$v=\frac{d_{tot}}{t_{tot}}$

the distance travelled by the trucker in the first 3 hour can be written as the time multiplied by the velocity:

$d_1 = (3 h)(60 mph)=180 mi$

So the total distance is

$d_{tot}=d_1 +d_2 = 180 mi+20 mi=200 mi$

The total time is equal to the first 3 hours + the time taken to cover the following 20 miles in the city:

$t_{tot}=3 h +t_2$

So, the equation can be rewritten as:

$v=\frac{d_{tot}}{t_{tot}}=\frac{(3 h)(60 mph)+20 mi}{3 h +t_2}$

b) 0.50 h (half a hour)

Since we know the value of the average speed, $v=57.14 mph$ , we can substitute it into the previous equation to find the value of $t_2$ , the time the trucker drove in the city:

$v=\frac{200 mi}{3h +t_2}\\3h+t_2 = \frac{200 mi}{v}\\t_2 = \frac{200 mi}{v}-3h=\frac{200 mi}{57.14 mph}-3 h=0.50 h$

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3 years ago

HELP TRIGONOMETRY!! Solve for x and y

zysi [14]

Use one of the many Trigonometric ratios to solve for the missing angle and then use Pythagorean theorem to solve for the missing side.

These things can be done in any order.

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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

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

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

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