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gtnhenbr [62]
3 years ago
8

8(10−k)=2k solve for k

Mathematics
2 answers:
Olenka [21]3 years ago
6 0
Plug in 8 for k and that's your answer
Delvig [45]3 years ago
3 0
<span>8(10−k)=2k
80 - 8k = 2k
80 = 10k
8 = k
or 
k = 8</span>
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Leni [432]

Answer: c - 6 = 20

Explanation:

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3 years ago
What percent of 15y is 75?
Firlakuza [10]

Answer:

20%

Step-by-step explanation:

because 15/75×100 =1/5×100=20%

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3 years ago
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The prior probabilities for events A1 and A2 are P(A1) = 0.30 and P(A2) = 0.55. It is also known that P(A1 ∩ A2) = 0. Suppose P(
k0ka [10]

We're given the following probabilities:

P(A_1)=0.30

P(A_2)=0.55

P(A_1\cap A_2)=0

P(B\mid A_1)=0.20

P(B\mid A_2)=0.05

(a) Yes, A_1 and A_2 are mutually exclusive. This is exactly what zero probability of their intersection means. The two events cannot occur simultaneously.

(b) Use the definition of conditional probability to expand:

P(A_1\cap B)=P(A_1)P(B\mid A_1)=0.30\cdot0.20=0.06

P(A_2\cap B)=P(A_2)P(B\mid A_2)=0.55\cdot0.05=0.0275

(c) By the law of total probability,

P(B)=P(A_1\cap B)+P(A_2\cap B)=0.06+0.0275=0.0875

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P(A_1\mid B)=\dfrac{P(A_1)P(B\mid A_1)}{P(B)}=\dfrac{0.30\cdot0.20}{0.0875}\approx0.686

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3 0
3 years ago
You must design a closed rectangular box of width w, length l and height h, whose volume is 504 cm3. The sides of the box cost 3
Charra [1.4K]

Answer:

Dimensions will be

Length = 7.23 cm

Width = 7.23 cm

Height = 9.64 cm

Step-by-step explanation:

A closed box has length = l cm

width of the box = w cm

height of the box = h cm

Volume of the rectangular box = lwh

504 = lwh

h=\frac{504}{lw}

Sides which involve length and width and height, cost = 3 cents per cm²

Top and bottom of the box costs = 4 cents per cm²

Cost of the sides C_{s}= 3[2(l + w)h] = 6(l + w)h

C_{s}= 3[2(l + w)h]

C_{s}=6(l+w)(\frac{504}{lw} )

Cost of the top and the bottom C_{(t,p)}= 4(2lw) = 8lw

Total cost of the box C = 3024\frac{(l+w)}{lw} + 8lw

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To minimize the cost of the sides

\frac{dC}{dl}=3024(-l^{-2}+0)+8w=0

\frac{3024}{l^{2}}=8w

\frac{378}{l^{2}}=w ---------(1)

\frac{dC}{dw}=3024(-w^{-2})+8l=0

\frac{3024}{w^{2}}=8l

\frac{378}{w^{2}}=l

w^{2}=\frac{378}{l}-------(2)

Now place the value of w from equation (1) to equation (2)

(\frac{378}{l^{2}})^{2}=\frac{378}{l}

\frac{(378)^{2} }{l^{4}}=\frac{378}{l}

l³ = 378

l = ∛378 = 7.23 cm

From equation (2)

w^{2}=\frac{378}{7.23}

w^{2}=52.28

w = 7.23 cm

As lwh = 504 cm³

(7.23)²h = 504

h=\frac{504}{(7.23)^{2}}

h = 9.64 cm                        

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3 years ago
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raketka [301]

Answer:

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1+2(x+2)

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