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Naddika [18.5K]
3 years ago
10

6.4 meters + 14.7 - 23.2 = ?

Mathematics
2 answers:
luda_lava [24]3 years ago
5 0

Answer:

-2.1

Step-by-step explanation:

hram777 [196]3 years ago
3 0

Answer:

-2.1

Step-by-step explanation:

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What does this equal? 3y=4
Nutka1998 [239]
3y = 4

First, take the 3 and divide both sides by it.
y =  \frac{4}{3}

Answer as fraction: \frac{4}{3}
Answer as decimal: 1.3333

4 0
3 years ago
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What is the solution for this system of equations?<br>x-6y+4z=2<br>2x+4y-8z=16<br>x-2y=5
Serga [27]
2(x-6y+4z=2)
2x+4y-8z=16

2x-12y+8z=4
2x+4y-8z=16

4x-8y=20
x-2y=5

4x-8y=20
-4x+8y=-20

infinite many solutions
7 0
2 years ago
I need help with this
Svet_ta [14]

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5 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
Many states assess the skills of their students in various grades. One program that is available for this purpose is the Nationa
Nataly [62]

Answer:

A score of 314 is needed to be in the top 25% of students who take this exam.

Step-by-step explanation:

We are given that one of the tests provided by the NAEP assesses the reading skills of twelfth-grade students. In recent years, the national mean score was 289 and the standard deviation was 37.

Let X = <u><em>scores of the tests provided by the NAEP</em></u>

The z-score probability distribution for the normal distribution is given by;

                           Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = national mean score = 289

           \sigma = standard deviation = 37

Now, we have to find how high a score is needed to be in the top 25% of students who take this exam, that means;

            P(X > x) = 0.25     {where x is the required score}

            P( \frac{X-\mu}{\sigma} > \frac{x-289}{37} ) = 0.25

            P(Z > \frac{x-289}{37} ) = 0.25

In the z table, the critial value of z that represents the top 25% of the area is given as 0.6745, that is;

                        \frac{x-289}{37}=0.6745

                        {x-289}=0.6745\times 37

                        x = 289 + 24.96 = 313.96 ≈ 314

Hence, a score of 314 is needed to be in the top 25% of students who take this exam.

7 0
3 years ago
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