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3 years ago

Natalie accelerator skateboard along a straight path for 4 milliseconds to 0 milliseconds in 25.0 seconds find the acceleration.

Mathematics

1 answer:

Katarina [22]3 years ago

8 0

M/s is meters per second not milliseconds :)
m/s^2 is the accelaration in m/s per second
v=v2-v1=0-4=-4
a=v/t
a=-4/25 m/s^2=-0.16 m/s^2

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Can someone solve this for me???

kolezko [41]

Answer:

Part A: x^2 + 5x - 8

Part B: 5x^3 + 3x^2 - 6x + 5

Step-by-step explanation:

Part A:

To find the total of side 1 and 2, add the two expressions which represent each side. Combine like terms to simplify.

3x^2 − 2x − 5 + (7x - 2x^2 − 3)

3x^2 - 2x^2 -2x + 7x -5 + -3

x^2 + 5x - 8

Part B:

To solve for the third side of a triangle using its perimeter, subtract the lengths of two of its sides from its perimeter. The perimeter is 5x^3 + 4x^2 − x − 3. Subtract 3x^2 − 2x − 5 and 7x - 2x^2 − 3. The remaining expression is the third side. Combine like terms to simplify.

5x^3 + 4x^2 − x − 3 - (3x^2 − 2x − 5) - (7x - 2x^2 − 3)

5x^3 + 4x^2 - x - 3 - 3x^2 + 2x + 5 - 7x + 2x^2 + 3

5x^3 + (4x^2 -3x^2 + 2x^2) + (-x + 2x - 7x) + (-3 + 5 + 3)

5x^3 + 3x^2 - 6x + 5

4 0

3 years ago

Which represents the inverse of the function f(x) = 4x?

Cloud [144]

To find the inverse function simply replace "x" by "y", replace "y" by "x", and solve for y
f(x)=4x
y=4x
x=4y
y=x/4 or h(x)=(1/4)x

8 0

1 year ago

Liam and Kaiden are brothers who go to different schools. The school day is just as long at Liam's school as at Kaiden's school.

Triss [41]

Answer:

He's not right

Step-by-step explanation:

L = 4/6 = 2/3

K = 2/3

4 0

2 years ago

Please help and number the questions<br> Giving 50 points please solve it<br> And I won’t report

maxonik [38]

Answer:

1. 18.84 in

2. 56.52 cm

3. 4.71 ft

4. 25.12 m

5. 37.68 ft

6. 12.56 yd

7. 43.96 in

8. 28.26 cm

9. 7.85 m

Step-by-step explanation:

$C=2\pi r$

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