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1 year ago

in the number 85,549 the 5 in the hundreds place is. times that of the 5 in the thousands place.

Mathematics

1 answer:

Oksanka [162]1 year ago

8 0

The value of 5 in the hundreds place is 500

The value of 5 in the thousands place is 5000

To find 500 is how many times in 5000 we will divide them

$\frac{500}{5000}=\frac{1}{10}$

Then 500 is 1/10 times that 5000

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What two numbers can add to 4 but multiple to -12

olasank [31]

Answer:

<em>Numbers: 6 and -2</em>

Step-by-step explanation:

<u>Equations</u>

This question can be solved by inspection. It's just a matter of factoring 12 into two factors that sum 4. Both numbers must be of different signs and they are 6 and -2. Their sum is indeed 6-2=4 and their product is 6*(-2)=-12.

However, we'll solve it by the use of equations. Let's call x and y to the numbers. They must comply:

$x+y=4\qquad\qquad [1]$

$x.y=-12\qquad\qquad [2]$

Solving [1] for y:

$y=4-x$

Substituting in [2]

$x(4-x)=-12$

Operating:

$4x-x^2=-12$

Rearranging:

$x^2-4x-12=0$

Solving with the quadratic formula:

$\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

With a=1, b=-4, c=-12:

$\displaystyle x=\frac{-(-4)\pm \sqrt{(-4)^2-4(1)(-12)}}{2(1)}$

$\displaystyle x=\frac{4\pm \sqrt{16+48}}{2}$

$\displaystyle x=\frac{4\pm 8}{2}$

The solutions are:

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7 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

Somebody please help

DENIUS [597]

What you want to do, is isolate the y by itself to rewrite the equation as y=mx+b. So when you move the 5x over the equation looks like this -6y=-5x+30. Then you divide the -5x+30 by -6 to isolate the y. Thus making the equation y=5/6x-5. And since there is only one answer with the y-intercept as (0,-5); The answer is A.

8 0

3 years ago