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DanielleElmas [232]

2 years ago

9

A store manager would like to set the three digit code. How many possible three digit code choices does he have if he can pick d

igits from 0 to 9 with repetition allowed?

1 answer:

Shalnov [3]2 years ago

6 0

10 numbers for each digit

so 10 x 10 x 10 = 1000

You might be interested in

What is the proof the formula for cos(A+B)=cosAcosB-sinAsinB?

brilliants [131]

Given data, cos(A - B) = cosAcosB + sinAsinB
<span>let, A=60' and B=30' ( here the ' sign bears degree) </span>
<span>L.H.S = cos(A - B) </span>
<span>=cos (60'-30) ( using value of A and B ) </span>
<span>=cos30' </span>
<span>= sqrt3/2 </span>
<span>R.H.S= cosAcosB + sinAsinB </span>
<span>=cos60' cos30' + sin60' sin30' </span>
<span>= 1/2* sqrt3/2+ sqrt3/2*1/2 </span>
<span>= sqrt3/4 + sqrt3/4 </span>
<span>=2 sqrt3/4 </span>
<span>= sqrt3/2 </span>
<span>so L.H.S =R.H.S or cos(A - B) = cosAcosB + sinAsinB</span>

8 0

2 years ago

A football field is 120 yards long by 53 yards wide if a player runs diagonally from one corner to the opposite corner. How far

qaws [65]

Answer: They will travel about 131.18 yards.

Step-by-step explanation:

Given : A football field is 120 yards long by 53 yards wide.

We know that a football field is rectangular in shape.

Each interior angle in a rectangle is a right angle.

Then, by Pythagoras theorem, we have

(Diagonal)² = (Length)² + (Width)²

If a player runs diagonally from one corner to the opposite corner, then the length of the diagonal is given by :-

$(\text{Diagonal})=\sqrt{(120)^2+(53)^2}\\\\\Rightarrow\ (\text{Diagonal})=\sqrt{14400+2809}\\\\\Rightarrow\ (\text{Diagonal})=\sqrt{17209}\\\\\Rightarrow\ (\text{Diagonal})=131.183078177\approx131.18\text{ yards}$

Hence, they will travel about 131.18 yards.

7 0

2 years ago

In 1898 L. J. Bortkiewicz published a book entitled The Law of Small Numbers. He used data collected over 20 years to show that

attashe74 [19]

Answer:

(a) The probability of more than one death in a corps in a year is 0.1252.

(b) The probability of no deaths in a corps over 7 years is 0.0130.

Step-by-step explanation:

Let <em>X</em> = number of soldiers killed by horse kicks in 1 year.

The random variable $X\sim Poisson(\lambda = 0.62)$ .

The probability function of a Poisson distribution is:

$P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0,1,2,...$

(a)

Compute the probability of more than one death in a corps in a year as follows:

P (X > 1) = 1 - P (X ≤ 1)

= 1 - P (X = 0) - P (X = 1)

$=1-\frac{e^{-0.62}(0.62)^{0}}{0!}-\frac{e^{-0.62}(0.62)^{1}}{1!}\\=1-0.54335-0.33144\\=0.12521\\\approx0.1252$

Thus, the probability of more than one death in a corps in a year is 0.1252.

(b)

The average deaths over 7 year period is: $\lambda=7\times0.62=4.34$

Compute the probability of no deaths in a corps over 7 years as follows:

$P(X=0)=\frac{e^{-4.34}(4.34)^{0}}{0!}=0.01304\approx0.0130$

Thus, the probability of no deaths in a corps over 7 years is 0.0130.

6 0

1 year ago

A school district has a student population of $6,734 students. If the maximum class size is 25 students. How many more teacher m

emmasim [6.3K]

Answer:

53 teachers

Step-by-step explanation:

Basically, what we need to do here is to find how many teachers there need to be, first. If there are 6,734 students in the school district and if maximum class size is 25, then the number of teachers needed is:

6,734 / 25 = 269.36

Of course, it's obvious that we can't have a decimal number of teachers, so we need to find integer (269 or 270).

If we take 269 teachers and 25 students per class, we get:

269 • 25 = 6,725 students, which is not enough, since there are 6,734 students.

That means that the number of teachers needed is 270.

It is given that there are already 217 teachers, meaning that 270-217=53 teachers have to be supplemented.

8 0

2 years ago

An entrepreneur estimates his total profit (total revenue minus total cost) for his proposed company as p(x) = x3 − 4x2 + 5x − 2

RideAnS [48]

Answer:

Year in which the entrepreneur break even is 4

Step-by-step explanation:

We are given:

p(x) = x^3-4x^2+5x-20

We would find the value of x

Solving:

x^3-4x^2+5x-20 = 0

(x^3-4x^2)+(5x-20) = 0

x^2(x-4)+5(x-4) = 0

(x-4)(x^2+5)=0

=> x-4 = 0 and x^2+5 =0

x = 4 and x^2 = -5

x = 4 and x = ±√-5 or ±√5 i (not real solutions)

So, x = 4

So, year in which the entrepreneur break even is 4

4 0

2 years ago