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creativ13 [48]
3 years ago
10

A "b" value of less than 1 produces a graph with exponential decay

Mathematics
1 answer:
Andru [333]3 years ago
5 0
<span>If this statement is a True or False question.

Then, TRUE. A "b" value of less than 1 produces a graph with exponential decay.

We consider this formula: y = a*b^x or f(x) =a*b^x

a is the initial value, b is the rate, x is the exponent. 

If b has a value of less than 1, as the exponent increases the resulting value of y decreases.

f(x) = 20(0.90)^x

f(1) = 200(0.90)</span>¹ = 180
f(2) = 200(0.90)² = 162
f(3) = 200(0.90)³ = 145.80
f(4) = 200(0.90)⁴ = 131.22
f(5) = 200(0.90)⁵ = 118.10

Exponential decay is the decrease in a quantity. As you can see in the given example the value of f(x) decreases as the value of x increases because the value of b is less than 1.
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Look at triangle ABC.
sashaice [31]

Answer:

20 units or 4.8

hope this helps!

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
The quality-control manager at a compact flourescent light bulb factory wants to test the claim that the mean life of a large sh
MAXImum [283]

Answer:

a. At the 0.05 level of significance,  there is evidence that the mean life is different from 6,500 hours.

b. The p value= ≈ 0.00480 for z- test which is less than 0.05 and H0 is rejected .

The p value= 0.006913 for t- test which is less than 0.05 and H0 is rejected for 49 degrees of freedom.

c. CI [6583.336 ,6816.336]

d.  The range of CI [6583.336 ,6816.336] tells that the cfls having a different mean life lie in this range.

Step-by-step explanation:

Population mean = u= 6500 hours.

Population standard deviation = σ=500 hours.

Sample size =n= 50

Sample mean =x`=  6,700 hours

Sample standard deviation=s=  600 hours.

Critical values, where P(Z > Z) =∝ and P(t >) =∝

Z(0.10)=1.282  

Z(0.05)=1.645  

Z(0.025)=1.960  

t(0.01)(49)= 1.299

t(0.05)= 1.677  

t(0.025,49)=2.010

Let the null and alternate hypotheses be

H0: u = 6500 against the claim Ha: u ≠ 6500

Applying Z test

Z= x`- u/ s/√n

z= 6700-6500/500/√50

Z= 200/70.7113

z= 2.82=2.82

Applying  t test

t= x`- u /s/√n

t= 6700-6500/600/√50

t= 2.82

a. At the 0.05 level of significance,  there is evidence that the mean life is different from 6,500 hours.

Yes we reject H0  for z- test as it falls in the critical region,at the 0.05 level of significance, z=2.82 > z∝=1.645

For t test  we reject H0   as it falls in the critical region,at the 0.05 level of significance, t=2.82 > t∝=1.677 with n-1 = 50-1 = 49 degrees of freedom.

b. The p value= ≈ 0.00480 for z- test which is less than 0.05 and H0 is rejected .

The p value= 0.006913 for t- test which is less than 0.05 and H0 is rejected for 49 degrees of freedom.

c. The 95 % confidence interval of the population mean life is estimated by

x` ±  z∝/2  (σ/√n )

6700± 1.645 (500/√50)

6700±116.336

6583.336 ,6816.336

d. The range of CI [6583.336 ,6816.336] tells that the cfls having a different mean life lie in this range.

6 0
3 years ago
Hi, can someone help on this?
marissa [1.9K]

Answer:

i beleive its 9 and 1 if im reading that right

Step-by-step explanation:

8 0
3 years ago
HELP I'M BEING TIMED FOR THIS TEST!!!
ICE Princess25 [194]
The correct answer is C
You must apply the power to both 5 and t
7 0
3 years ago
A.Find a formula for
Vlada [557]

a. Notice that

1/(1*2) = 1/2 = 1 - 1/2

1/(2*3) = 1/6 = 1/2 - 1/3

1/(3*4) = 1/12 = 1/3 - 1/4

and so on, which suggests the n-th term of the sum can be written as

\dfrac1{n(n+1)}=\dfrac1n-\dfrac1{n+1}

Then the sum itself is telescoping:

\dfrac1{1\cdot2}+\dfrac1{2\cdot3}+\dfrac1{3\cdot4}+\cdots+\dfrac1{n(n+1)}

=\left(1-\dfrac12\right)+\left(\dfrac12-\dfrac13\right)+\left(\dfrac13-\dfrac14\right)+\cdots+\left(\dfrac1n-\dfrac1{n+1}\right)

=1-\dfrac1{n+1}

b. The proof is trivial:

\dfrac1n-\dfrac1{n+1}=\dfrac{n+1}{n(n+1)}-\dfrac n{n(n+1)}=\dfrac{n+1-n}{n(n+1)}=\dfrac n{n+1}

so the formula found in (a) is correct.

5 0
3 years ago
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