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3 years ago

PLS help ASAP this really urgent

Mathematics

2 answers:

igor_vitrenko [27]3 years ago

8 0

Skewed right. I just had this question.

MrMuchimi3 years ago

7 0

Answer:

Skewed right.

Step-by-step explanation:

The graph is not symmetric because it's not an exact mirror of each side if you drew a line in the center. If the graph has highpoints on the left side and is leaning down to the right, it's skewed right. Same thing vice versa.

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What is the height of an object thrown when time is 6 seconds? Also explain how u got your answer

nevsk [136]

Answer:

$946.4\ m$

Step-by-step explanation:

Let

h(t) ---> the height of an object in meters

t ----> the time in seconds

we have

$h(t)=-2.6t^2+40t+800$

For t=6 sec

substitute in the quadratic equation the value of t and solve for h

$h(6)=-2.6(6)^2+40(6)+800$

$h(6)=946.4\ m$

4 0

3 years ago

Help plzzzzzz plz help this ? is realy hard Simplify the expression 7(a−3+4b).

velikii [3]

7(a−3+4b)

distribute

7a-7*3+7*4b

7a-21+28b

4 0

3 years ago

Read 2 more answers

12. What is the value of x ?

Firlakuza [10]

Answer:

If that dot in the middle is the middle of the circle, then that should mean the center is 5 away from both of those lines. Therefore, I believe x would 9/2 or 4.5

4 0

3 years ago

Simplify

jeka57 [31]

$\bf ~\hspace{7em}\textit{negative exponents} \\\\ a^{-n} \implies \cfrac{1}{a^n} ~\hspace{4.5em} a^n\implies \cfrac{1}{a^{-n}} ~\hspace{4.5em} \cfrac{a^n}{a^m}\implies a^na^{-m}\implies a^{n-m} \\\\[-0.35em] ~\dotfill\\\\ x^5\div y^2\times x^3\times y^5\implies \cfrac{x^5}{y^2}\cdot x^3\cdot y^5\implies x^5\cdot y^{-2}\cdot x^3\cdot y^5 \\\\\\ x^{5+3}y^{-2+5}\implies x^8y^3$

4 0

3 years ago

Read 2 more answers

Suppose that – in any given time period – a certain stock is equally likely to go up 1 unit or down 1 unit, and that the outcome

Scorpion4ik [409]

Answer:

Step-by-step explanation:

From the given information:

Let X represent the amount of the stock that goes up in the first period; &

Let Y denote the cumulative amount that goes up in the first three periods.

Then,

$X= X_1 + X_2 \\ \\ \\EX= EX_1 +EX_2\\ \\\\Now ; EX = 0 \\ \\ EX = \dfrac{1}{2}\times 1 + \dfrac{1}{2}\times -1 \\ \\ EX = \dfrac{1}{2}-\dfrac{1}{2} \\ \\ EX = 0$

$Cov(x,y) = Var(X_1) +Var(X_2) \\ \\ =\dfrac{1}{2}(1+1) + \dfrac{1}{2}(1+1) \\ \\ = \dfrac{1}{2}(2)+ \dfrac{1}{2}(2) \\ \\ = 1+1 \\ \\ =2$

$Var (X) = Var (X_1+X_2)\\ \\ = 2 \ Var(X_1) \\ \\ = 2 \times (1) \\ \\ = 2$

$Var (Y) = Var (X_1 +X_2+X_3) \\ \\ = 3 \times Var(X_1) = \\ \\ = 3 \times 1\\ \\ = 3$

∴

$Corr(x,y) = \dfrac{2}{\sqrt{6}} \\ \\ = \dfrac{2}{\sqrt{2\times 3}} \\ \\= \sqrt{\dfrac{2}{3}}$

3 0

3 years ago