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3 years ago

Which side is the included side for JKL and KLJ ? JK KL JL

Mathematics

2 answers:

qaws [65]3 years ago

8 0

Answer is KL
hope it helps

Aneli [31]3 years ago

8 0

Its KL because both of them have that side

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Can someone please give me the (Answers) to this? ... please ...<br><br> I need help….

Rina8888 [55]

Step-by-step explanation:

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2 years ago

ILL MARK BRAINIEST IF YOU DO THIS RIGHT!!! DO 9 AND 10!!!

yuradex [85]

Number 9 is L is greater than or equal to 110

4 0

3 years ago

NEEDED BY TOMORROW!

marissa [1.9K]

Answer:

$0.375$ , Terminating

Step-by-step explanation:

Mackenzie has 32 books and 12 of which are by her favorite author.

One of the easier ways to convert something to a decimal is to first start as a fraction. We need to have it be the number of books by her favorite author (12) out of the total number of books that she owns (32)

This in the form of a fraction would be $\frac{12}{32}$

Now, let us reduce this fraction.

Both terms can be divided by 4

$\frac{12}{32}/\frac{4}{4} \\\\\frac{3}{8}$

Recall that $\frac{1}{8} =0.125$

This means that

$\frac{3}{8} =3*0.125=0.375$

As you can see. this decimal stops after 3 places, which means that it is terminating.

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

PLEASE HELP IM STUCK

mrs_skeptik [129]

The number of outcomes possible from flipping each coin is 2, therefore;

The expression that can be used to find the number of outcomes for flipping 4 coins is: 2•2•2•2

<h3>How can the expression for the number of combinations be found?</h3>

The possible outcome of flipping 4 coins is given by the sum of the possible combinations of outcomes as follows;

The number of possible outcome from flipping the first coin = 2 (heads or tails)

The outcomes from flipping the second coin = 2

The outcome from flipping the third coin = 2

The outcome from flipping the fourth coin = 2

The combined outcome is therefore;

Outcome from flipping the 4 coins = 2 × 2 × 2 × 2

The correct option is therefore;

2•2•2•2

Learn more about finding the number of combinations of items here:

brainly.com/question/4658834

#SPJ1

7 0

2 years ago