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Andrews [41]
1 year ago
13

RS=2x-8, ST= 11, RT= x+10

Mathematics
1 answer:
swat321 year ago
4 0

Answer:

X=7 ST=11 RT=17

Step-by-step explanation:

RT=RS+ST. RS= 2(7)-8. ST=11

X+10=2x-8+11. =14-8=6. RT= x+10

X+10=2x+3. RS=6. 7+10=17

10=x+3. RT=17

X=7

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Butoxors [25]
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1 year ago
3. After completing a statistical analysis of a survey of 40 students, the principal
Serhud [2]

Answer:

Option A

Step-by-step explanation:

A type I error is committed when a researcher rejects the null hypothesis when it is actually true.

The null hypothesis is: U <= 50%

The alternative is: U > 50%

Thus, the principal could have committed an error by rejecting null hypothesis and concluding that more than 50% of students want earlier lunch, when in actuality 50% or less want earlier lunch.

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1 year ago
Show that the Fibonacci numbers satisfy the recurrence relation fn = 5fn−4 + 3fn−5 for n = 5, 6, 7, . . . , together with the in
Sonja [21]

Answer with step-by-step explanation:

We are given that the recurrence relation

f_n=5f_{n-4}+3f_{n-5}

for n=5,6,7,..

Initial condition

f_0=0,f_1=1,f_2=1,f_3=2,f_4=3

We have to show that Fibonacci numbers satisfies the recurrence relation.

The recurrence relation of Fibonacci numbers

f_n=f_{n-1}+f_{n-2},f_0=0,f_1=1

Apply this

f_n=(f_{n-2}+f_{n-3})+f_{n-2}=2f_{n-2}+f_{n-3}

f_n=2(f_{n-3}+f_{n-4})+f_{n-3}=3f_{n-3}+2f_{n-4}

f_n=3(f_{n-4}+f_{n-5})+2f_{n-4}=5f_{n-4}+3f_{n-5}

Substitute n=2

f_2=f_1+f_0=1+0=1

f_3=f_2+f_1=1+1=2

f_4=f_3+f_2=2+1=3

Hence, the Fibonacci numbers satisfied the given recurrence relation .

Now, we have to show that f_{5n} is divisible by 5 for n=1,2,3,..

Now replace n by 5n

f_{5n}=5f_{5n-4}+3f_{5n-5}

Apply induction

Substitute n=1

f_5=5f_1+3f_0=5+0=5

It is true for n=1

Suppose it is true for n=k

f_{5k}=5f_{5k-4}+3f_{5k-5} is divisible 5

Let f_{5k}=5q

Now, we shall prove that for n=k+1 is true

f_{5k+5}=5f_{5k+5-4}+3f_{5k+5-5}=5f_{5k+1}+3f_{5k}=5f_{5k+1}+3(5q)

f_{5k+5}=5(f_{5k+1}+3q)

It is multiple of 5 .Therefore, it is divisible by 5.

It is true for n=k+1

Hence, the f_{5n} is divisible by 5 for n=1,2,3,..

8 0
1 year ago
The area of the regular hexagon is 169.74 ft2. A regular hexagon has an apothem with length 7 feet and an area of 169.74 feet sq
lorasvet [3.4K]

Answer:

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Step-by-step explanation:

In this question, we are asked to calculate the perimeter of a regular hexagon, given the length of its apothem and its area value

Recall;

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Plugging these values into the equation, we have;

169.74 = 0.5 * 7 * P

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8 0
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