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2 years ago

Identify the initial condition and the rate of change y=−5x+32

Mathematics

2 answers:

ss7ja [257]2 years ago

5 0

The number next to x is rate of change, so in this case -5, and the number thats alone is the initial so 32. So ans A is correct.

insens350 [35]2 years ago

3 0

for the given graph :

y = -5x + 32

rate of change = -5

Initial condition = 32

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2 years ago

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A jacket with an original price of $98.60 is discounted 65%. What is the Sale Price?

fiasKO [112]

Answer:

34.51

Step-by-step explanation:

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3 years ago

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Find the general term of {a_n}

Assoli18 [71]

From the given recurrence, it follows that

$a_{n+1} = 2a_n + 1 \\\\ a_{n+1} = 2(2a_{n-1} + 1) = 2^2a_{n-1} + 1 + 2 \\\\ a_{n+1} = 2^2(2a_{n-2}+1) + 1 + 2 = 2^3a_{n-2} + 1 + 2 + 2^2 \\\\ a_{n+1} = 2^3(2a_{n-3} + 1) + 1 + 2 + 2^2 = 2^4a_{n-3} + 1 + 2 + 2^2 + 2^3$

and so on down to the first term,

$a_{n+1} = 2^na_1 + \displaystyle \sum_{k=0}^{n-1}2^k$

(Notice how the exponent on the 2 and the subscript of a in the first term add up to n + 1.)

Denote the remaining sum by S ; then

$S = 1 + 2 + 2^2 + \cdots + 2^{n-1}$

Multiply both sides by 2 :

$2S = 2 + 2^2 + 2^3 + \cdots + 2^n$

Subtract 2S from S to get

$S - 2S = 1 - 2^n \implies S = 2^n - 1$

So, we end up with

$a_{n+1} = 4\cdot2^n + S \\\\ a_{n+1} = 2^2\cdot2^n + 2^n-1 \\\\ a_{n+1} = 2^{n+2} + 2^n - 1 \\\\\implies \boxed{a_n = 2^{n+1} + 2^{n-1} - 1}$

5 0

2 years ago

7652 divided by 12.3

Lady_Fox [76]

Answer:

The answer to the question provided is 622.1138211.

Rounded to the nearest tenth is 622.1

3 0

2 years ago

A sample of 15 commuters in Chicago showed the average of the commuting times was 33.2 minutes. If s = 8.3 minutes, find the 95%

OleMash [197]

Answer:

The 95% confidence interval of the true mean.

(29.4261 ,36.9739)

Step-by-step explanation:

Step :- (i)

Given sample size 'n' =15

sample of the mean x⁻ = 33.2

The standard deviation of the sample 'S' = 8.3

95% of confidence intervals

$(x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } ,x^{-} + t_{\alpha }\frac{S}{\sqrt{n} } )$

Step:-(ii)

The degrees of freedom γ=n-1 = 15-1=14

The tabulated value t = 1.761 at 0.05 level of significance.

now substitute all possible values, we get

$(33.2 - 1.761\frac{8.3}{\sqrt{15} } ,33.2+ 1.761\frac{8}{\sqrt{15} } )$

After calculation , we get

(33.2-3.7739 , 33.2+3.7739

(29.4261 ,36.9739)

Conclusion:-

the 95% confidence interval of the true mean.

(29.4261 ,36.9739)

8 0

3 years ago