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4 years ago

Questions 1-6 please

Mathematics

1 answer:

MakcuM [25]4 years ago

6 0

Question 1 is $6 \sqrt{2}$
question 2 is $4 \sqrt{5} + 3 \sqrt{2}$
question 3 is $6 \sqrt{2}$
question 4 is $8 \sqrt{3}$
question 5 is $\frac{ \sqrt{3}}{2}- \sqrt{3}$
question 6 is $5$

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The probability distribution of the amount of memory X (GB) in a purchased flash drive is given below. x 1 2 4 8 16 p(x) .05 .10

Rama09 [41]

Answer:

a) $E(X) = 6.45$

b) $E(X^{2} )= 57.25$

c) $V(X) = 15.648$

d) E(3X + 2) = 21.35

e) $E(3X^{2} +2) = 173.75$

f) V(3X+2) = 140.832

g) E(X+1) = 7.45

h) V(X+1) = 15.648

Step-by-step explanation:

a) $E(X) = \sum xP(x)$

$E(X) = (1*0.05) + (2*0.10) + (4*0.35) + (8*0.40) + (16*0.10)\\E(X) = 6.45$

$E(X^{2} ) = (1^{2} *0.05) + (2^{2} *0.10) + (4^{2} *0.35) + (8^{2} *0.40) + (16^{2} *0.10)\\ E(X^{2} )= 57.25$

$V(X) = E(X^{2} ) - (E(X))^{2} \\V(X) = 57.25 - 6.45^{2} \\V(X) = 15.648$

$E(3X+2) = 3E(X) + 2\\E(3X+2) = (3*6.45) + 2 \\E(3X+2) = 21.35$

$E(3X^{2} +2) = 3E(X^{2} ) + 2\\E(3X^{2} +2) = (3*57.25) + 2 \\E(3X^{2} +2) = 173.75$

$V(3X+2) = 3^{2} V(X)\\V(3X+2) = 9*15.648\\V(3X+2) = 140.832$

$E(X+1) = E(X) + 1\\E(X+1) = 6.45 + 1\\E(X+1) =7.45$

$V(X+1) = 1^{2} V(X)\\V(X+1) = 15.648$

6 0

3 years ago

Which is the largest number? 8.2, 7.98, 8 9/10, 7 8/9

zzz [600]

Answer:

8 9/10 is the largest number

8 0

3 years ago

Read 2 more answers

Estimate the difference. 9 1/2- 1 6/7

Sav [38]

Answer:

$= 7 \frac{9}{14} \\$

Step-by-step explanation:

$9 \frac{1}{2} - 1 \frac{6}{7} \\ \frac{19}{2} - \frac{13}{7} \\ \frac{133 - 26}{14} \\ = \frac{107}{14} \\ = 7 \frac{9}{14}$

hope this helps

brainliest appreciated

good luck! have a nice day!

6 0

3 years ago

A function g(x) decreases by 7 over every unit interval in x and g(0) = 0

Zina [86]

Are you asking what the answer is because you are not asking us anything. I would really like to help you out but I need to know what the question is first.

5 0

3 years ago

The complex number _______ lies in the shaded area of the graph. The complex number _______ does not lie in the shaded area.

Anastaziya [24]

Think of imaginary numbers on a complex plane as coordinates - take the real part, that's the x coordinate, the imaginary part that's the y coordinate.

Therefore, for the first blank, -1+3i lies in the shaded area, because its coordinates are (-1,3).
And following on from that for the second one, -4i does not lie in the shaded area, since its coordinates are (0,-4).

Tl;dr is taking a complex number a+bi and placing it on a complex plane, it will have coordinates (a,b)

5 0

3 years ago