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3 years ago

14

The cafeteria staff makes 100 ham sandwiches, 70 tuna sandwiches and 40 turkey sandwiches. The sandwiches are all wrapped in the

same
plain, white paper and mixed together. Unfortunately, the cafeteria staff failed to mark the sandwiches. What is the probability that you and
your two friends, who are following you in line, will each get a turkey sandwich?

A. 0.00691
B. 0.00640
C 0.00649
D. 0.00573

1 answer:

Rashid [163]3 years ago

3 0

Answer:

Ooof...I'd say C because it's a very small chance unless they check the things first.Also the math said sooo

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A pound of weed and feed covers 200 square feet how many pounds are needed to fertilize a lawn measuring 70 feet by 50 feet

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You would have to find out how many square feet the lawn is first:
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A rectangular box with a volume of 272ft^3 is to be constructed with a square base and top. The cost per square foot for the bot

ASHA 777 [7]

Answer:

The dimensions of the box is 3 ft by 3 ft by 30.22 ft.

The length of one side of the base of the given box is 3 ft.

The height of the box is 30.22 ft.

Step-by-step explanation:

Given that, a rectangular box with volume of 272 cubic ft.

Assume height of the box be h and the length of one side of the square base of the box is x.

Area of the base is = $(x\times x)$

$=x^2$

The volume of the box is = area of the base × height

$=x^2h$

Therefore,

$x^2h=272$

$\Rightarrow h=\frac{272}{x^2}$

The cost per square foot for bottom is 20 cent.

The cost to construct of the bottom of the box is

=area of the bottom ×20

$=20x^2$ cents

The cost per square foot for top is 10 cent.

The cost to construct of the top of the box is

=area of the top ×10

$=10x^2$ cents

The cost per square foot for side is 1.5 cent.

The cost to construct of the sides of the box is

=area of the side ×1.5

$=4xh\times 1.5$ cents

$=6xh$ cents

Total cost = $(20x^2+10x^2+6xh)$

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Let

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Putting the value of h

$C=30x^2+6x\times \frac{272}{x^2}$

$\Rightarrow C=30x^2+\frac{1632}{x}$

Differentiating with respect to x

$C'=60x-\frac{1632}{x^2}$

Again differentiating with respect to x

$C''=60+\frac{3264}{x^3}$

Now set C'=0

$60x-\frac{1632}{x^2}=0$

$\Rightarrow 60x=\frac{1632}{x^2}$

$\Rightarrow x^3=\frac{1632}{60}$

$\Rightarrow x\approx 3$

Now $C''|_{x=3}=60+\frac{3264}{3^3}>0$

Since at x=3 , C''>0. So at x=3, C has a minimum value.

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The dimensions of the box is 3 ft by 3 ft by 30.22 ft.

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Answer:

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Step-by-step explanation:

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2 years ago

Someone please help me with this!!!

Alexxx [7]

34.I
for this problem you just multiply the 20 by 2=40
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$\frac{43}{20}$

35.G
if you divide
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$\frac{58}{25}$

36. C
$\frac{5}{5} = 1$
so if you subtract
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will equal

$\frac{1}{5}$
$1 \frac{1}{5}$

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