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3 years ago

read the quote at the beginning of this section. interpret the quote in terms of what you have learned about probability

Mathematics

1 answer:

Tema [17]3 years ago

7 0

Where’s the quote ? You didn’t provide it

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5. ALGEBRA Refer to the figure in

statuscvo [17]

Answer:

the value of x equals to 20 thats is solution okay

8 0

3 years ago

At a hardware store a bag holds up to 1 pound of nails. Each penny nail (x) weighs 0.05 pounds and each wood nail (y) weighs 0.1

Snowcat [4.5K]

0.05x + 0.1y < = 1 (thats less then or equal)

3 0

3 years ago

Read 2 more answers

7 Tom claims that the system of equations at the right has

olganol [36]

If you could provide a picture of the graph that would be very helpful

4 0

3 years ago

Read 2 more answers

Need help please...answer correctly

Lelechka [254]

Answer:

9.09

Step-by-step explanation:

x=speed of boat

y=speed of water current

Downstream relative speed = x+y

Upstream relative speed = x-y

Distance remains the same for both upstream and downstream.

a) Distance travelled downstream = speed x time = 3(x+y)

Distance travelled upstream = speed x time = 3.6(x-y)

b) Since both distances are equal, we can write

3(x+y)= 3.6(x-y)

3x+3y = 3.6x-3.6y

6.6y = 0,6x

x=11y: y=x/11

c) Water current has speed as 1/11 times of that of boat

In percent this equals 100/11 = 9.09%

3 0

3 years ago

Read 2 more answers

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

= $3024[\frac{1}{l}+\frac{1}{w}]$ + 8lw

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

6 0

4 years ago