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Veseljchak [2.6K]
3 years ago
7

Kate walks half a mile to the library. how many yards does she walk?

Mathematics
2 answers:
saw5 [17]3 years ago
8 0
0.5 miles converted to yards= 880 yards
Novosadov [1.4K]3 years ago
5 0
880 yards.

Explanation:

In one mile, there are 1760 yards. To find half of that, divide 1760 by two.
1760 ÷ 2 = 880
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A certain type of digital camera comes in either a 3-megapixel version or a 4 megapixel version. A camera store has received a s
Morgarella [4.7K]

Answer:

a) Hyper-geometric distribution

b1) P(X=2) = 0.424

b2) P(X>2) = 0.152

C1) Mean of X = 1.67

C2) Standard Deviation of X = 0.65

Step-by-step explanation:

b1) P(X = 2) = (5C2 * 7C2)/(12C4) = (10 * 21)/(495) = 210/495

P(X=2) = 0.424

b2) P(X>2) = 1 - P(X≤2) = 1 - [P(X=0) + P(X=1) + P(X=2)]

P(X=0) = (7C4)/(12C4) = 35/495 = 0.071

P(X=1) =  (5C1 * 7C3)/(12C4) = (5 * 35)/(495) = 175/495 = 0.354

P(X=2) = 0.424

P(X>2) = 1 - P(X≤2) = 1 - [0.424 + 0.071 + 0.354]

P(X>2) = 0.152

C1) Mean of X = nk/N

k = number of 3-megapixel cameras = 5

n = number of selected cameras = 4

N = Total number of cameras = 12

Mean of X = 5*4/12

Mean of X = 1.67

C2) Standard Deviation of X

Standard deviation of X=(nk/N*(1-k/N)*(N-n)/(N-1))1/2

Standard Deviation of X=(4*(5/20)*(1-5/12)*((12-4)/(12-1))^1/2 =0.65

Standard Deviation of X = 0.65

4 0
3 years ago
Which equation has the same solution as 4-2 (1-5) = 1 – 19?
lesya [120]
The answer will be D
3 0
3 years ago
Find an integer x such that 0<=x<527 and x^37===3 mod 527
Greeley [361]
Since 527=17\times31, we have that

x^{37}\equiv3\mod{527}\implies\begin{cases}x^{37}\equiv3\mod{17}\\x^{37}\equiv3\mod{31}\end{cases}

By Fermat's little theorem, and the fact that 37=2(17)+3=1(31)+6, we know that

x^{37}\equiv(x^2)^{17}x^3\equiv x^5\mod{17}
x^{37}\equiv(x^1)^{31}x^6\equiv x^7\mod{31}

so we have

\begin{cases}x^5\equiv3\mod{17}\\x^7\equiv3\mod{31}\end{cases}

Consider the first case. By Fermat's little theorem, we know that

x^{17}\equiv x^{16}x\equiv x\mod{17}

so if we were to raise x^5 to the nth power such that

(x^5)^n\equiv x^{5n}\equiv x\mod{17}

we would need to choose n such that 5n\equiv1\mod{16} (because 16+1\equiv1\mod{16}). We can find such an n by applying the Euclidean algorithm:

16=3(5)+1
\implies1=16-3(5)
\implies16-3(5)\equiv-3(5)\equiv1\mod{16}

which makes -3\equiv13\mod{16} the inverse of 5 modulo 16, and so n=13.

Now,

x^5\equiv3\mod{17}
\implies (x^5)^{13}\equiv x^{65}\equiv x\equiv3^{13}\equiv(3^4)^2\times3^4\times3^1\mod{17}

3^1\equiv3\mod{17}
3^4\equiv81\equiv4(17)+13\equiv13\equiv-4\mod{17}
3^8\equiv(3^4)^2\equiv(-4)^2\mod{17}
\implies3^{13}\equiv(-4)^2\times(-4)\times3\equiv(-1)\times(-4)\times3\equiv12\mod{17}

Similarly, we can look for m such that 7m\equiv1\mod{30}. Apply the Euclidean algorithm:

30=4(7)+2
7=3(2)+1
\implies1=7-3(2)=7-3(30-4(7))=13(7)-3(30)
\implies13(7)-3(30)\equiv13(7)equiv1\mod{30}

so that m=13 is also the inverse of 7 modulo 30.

And similarly,

x^7\equiv3\mod{31}[/ex] [tex]\implies (x^7)^{13}\equiv3^{13}\mod{31}

Decomposing the power of 3 in a similar fashion, we have

3^{13}\equiv(3^3)^4\times3\mod{31}

3\equiv3\mod{31}
3^3\equiv27\equiv-4\mod{31}
\implies3^{13}\equiv(-4)^4\times3\equiv256\times3\equiv(8(31)+8)\times3\equiv24\mod{31}

So we have two linear congruences,

\begin{cases}x\equiv12\mod{17}\\x\equiv24\mod{31}\end{cases}

and because \mathrm{gcd}\,(17,31)=1, we can use the Chinese remainder theorem to solve for x.

Suppose x=31+17. Then modulo 17, we have

x\equiv31\equiv14\mod{17}

but we want to obtain x\equiv12\mod{17}. So let's assume x=31y+17, so that modulo 17 this reduces to

x\equiv31y+17\equiv14y\equiv1\mod{17}

Using the Euclidean algorithm:

17=1(14)+3
14=4(3)+2
3=1(2)+1
\implies1=3-2=5(3)-14=5(17)-6(14)
\implies-6(14)\equiv11(14)\equiv1\mod{17}

we find that y=11 is the inverse of 14 modulo 17, and so multiplying by 12, we guarantee that we are left with 12 modulo 17:

x\equiv31(11)(12)+17\equiv12\mod{17}

To satisfy the second condition that x\equiv24\mod{31}, taking x modulo 31 gives

x\equiv31(11)(12)+17\equiv17\mod{31}

To get this remainder to be 24, we first multiply by the inverse of 17 modulo 31, then multiply by 24. So let's find z such that 17z\equiv1\mod{31}. Euclidean algorithm:

31=1(17)+14
17=1(14)+3

and so on - we've already done this. So z=11 is the inverse of 17 modulo 31. Now, we take

x\equiv31(11)(12)+17(11)(24)\equiv24\mod{31}

as required. This means the congruence x^{37}\equiv3\mod{527} is satisfied by

x=31(11)(12)+17(11)(24)=8580

We want 0\le x, so just subtract as many multples of 527 from 8580 until this occurs.

8580=16(527)+148\implies x=148
3 0
3 years ago
Find the area of the triangle 4ft 1ft 8ft please answer need help ASAP thank you ☺️
strojnjashka [21]

Answer:

16

Step-by-step explanation:

3 0
3 years ago
Carolyn was training for cross-country season on an indoor track. During her
Ahat [919]

Answer:

2.64 km

Step-by-step explanation:

24 laps each of 110 meters means a total of :

24 * 110 = 2640 m

which translated to km are: 2.64 km

6 0
3 years ago
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