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3 years ago

I need help I suck with lines

Mathematics

1 answer:

densk [106]3 years ago

5 0

Answer:

Your answer is correct. What is it that you need help on?

Step-by-step explanation:

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A restaurant offers a $12 dinner special that has 77 choices for an appetizer, 1010 choices for an entrée, and 44 choices for

krok68 [10]

<span>280 I'm assuming that this question is badly formatted and that the actual number of appetizers is 7, the number of entres is 10, and that there's 4 choices of desserts. So let's take each course by itself. You can choose 1 of 7 appetizers. So we have n = 7 After that, you chose an entre, so the number of possible meals to this point is n = 7 * 10 = 70 Finally, you finish off with a dessert, so the number of meals is: n = 70 * 4 = 280 Therefore the number of possible meals you can have is 280. Note: If the values of 77, 1010 and 44 aren't errors, but are actually correct, then the number of meals is n = 77 * 1010 * 44 = 3421880 But I believe that it's highly unlikely that the numbers in this problem are correct. Just imagine the amount of time it would take for someone to read a menu with over a thousand entres in it. And working in that kitchen would be an absolute nightmare.</span>

5 0

3 years ago

10. There are nine golf balls numbered from 1 to 9 in a bag. Three balls are randomly selected without replacement to form a 3-d

lbvjy [14]

Answer:

a) 504

b) 56

c) 0.111

Step-by-step explanation:

Data provided in the question:

There are nine golf balls numbered from 1 to 9 in a bag

Three balls are randomly selected without replacement

a) 3-digit numbers that can be formed

= $^nP_r$

n = 9

r = 3

= ⁹P₃

= $\frac{9!}{9!-3!}$

= 9 × 8 × 7

= 504

b) 3-digit numbers start with the digit 1

= _ _ _

in the above 3 blanks first digit is fixed i.e 1

we and we have 8 choices left for the last 2 digits

Thus,

n = 8

r = 2

Therefore,

= 1 × ⁸P₂

= 1 × $\frac{8!}{8!-2!}$

= 1 × 8 × 7

= 56

c) Probability that the 3-digit number formed is less than 200

Now,

The number of 3-digit number formed is less than 200 will be the 3-digit numbers start with the digit 1 i.e part b)

and total 3-digit numbers that can be formed is part a)

therefore,

Probability that the 3-digit number formed is less than 200

= 56 ÷ 504

= 0.111

5 0

3 years ago

Hey there people Im doing a poll

quester [9]

My favorite is probably aweSAMdude

5 0

3 years ago

You are looking at the New York ball drop on New Year’s Eve at a distance of 100 m away from the base of the structure. If the b

Fofino [41]

The question is an illustration of related rates.

The rate of change between you and the ball is 0.01 rad per second

I added an attachment to illustrate the given parameters.

The representations on the attachment are:

$\mathbf{x = 100\ m}$

$\mathbf{\frac{dy}{dt} = 2\ ms^{-1}}$ ---- the rate

$\mathbf{\theta = \frac{\pi}{4}}$

First, we calculate the vertical distance (y) using tangent ratio

$\mathbf{\tan(\theta) = \frac{y}{x}}$

Substitute 100 for x

$\mathbf{y = 100\tan(\theta) }$

$\mathbf{\tan(\theta) = \frac{y}{100}}$

Differentiate both sides with respect to time (t)

$\mathbf{ \sec^2(\theta) \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot \frac{dy}{dt}}$

Substitute values for the rates and $\mathbf{\theta }$

$\mathbf{ \sec^2(\pi/4) \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot 2}$

This gives

$\mathbf{ (\sqrt 2)^2 \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot 2}$

$\mathbf{ 2 \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot 2}$

Divide both sides by 2

$\mathbf{ \frac{d\theta}{dt} = \frac{1}{100} }$

$\mathbf{ \frac{d\theta}{dt} = 0.01 }$

Hence, the rate of change between you and the ball is 0.01 rad per second