A 5 pound bag of clay cost $2.65. At this rate how much would a 2 pound bag cost?

Dustin bought a DVD player that was marked down from $150 to $100. A video game console was $250 and marked down by $75. Which d

Vika [28.1K]

Answer:

It would be the game console

Step-by-step explanation:

The DVD player went down $50 the console went down about $100 and that is a better idea

3 0

3 years ago

Read 2 more answers

choose the point-slope form of the equation below that represents the line that passes through the points (-3,2) and (2,1)

UkoKoshka [18]

The answer should be: (2 - 1) = m(-3 - 2)

4 0

2 years ago

Read 2 more answers

HELP please!

mars1129 [50]

The answer would be B, (6,-3) because when you plug in the x and y values into the inequality, it is the only result that is true

8 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

8) 4.8.12.5.10. -17.

ASHA 777 [7]

Answer:

1) 1 3 7 13 21 31 43 ( +2, +4, +6, +8, +10 и т.д.)

2) 255 127 63 31 15 7 3 (число -1, а затем полученное число :2)

3)92 46 44 22 20 10 8 4 ( :2, -2, :2, -2, :2, -2 и т.д.)

4) 1 2 4 7 28 33 198 (+1, *2, +3, *4, +5, *6, +7, *8 и т.д)

5)3 7 6 7 9 7 12 7 15 7 (первое, третье, пятое и т.д. числа +3; между ними 7)

6) 6 8 11 13 16 18 21 23 (+2, +3, +2, +3 и т.д.)

7)1 4 8 13 19 26 34 43 (+3, +4, +5, +6, +7, +8 и т.д.)

8)128 64 32 16 8 4 2 (:2, :2, :2, :2 и т.д.)

9)3 4 8 17 33 58 94 (+1, +4, +9, +16, +25, +36 и т.д. (т.е. то, что прибавляется-увеличивается на +3, +5, +7, +9, +11, и т.д.))

10)47 39 32 26 21 17 14 12 (-8, -7, -6, -5, -4, -3 и т.д.)

11)256 37 64 31 16 25 4 19 (первое, третье, пятое и т.д. числа :4; а второе, четвертое, шестое и т.д. числа -6)

12)5 9 13 17 21 25 29 (+4, +4, +4, +4, +4 и т.д.)

13)1 4 3 6 5 8 7 10 (+3, -1, +3, -1, +3, -1 и т.д.)

14)2 4 6 12 14 28 30 60 (*2, +2, *2, +2, *2, +2 и т.д.)

15)11 12 10 13 9 14 8 15 (+1, -2, +3, -4, +5, -6, +7, -8 и т.д.)

Step-by-step explanation:

8 0

2 years ago