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Butoxors [25]
3 years ago
12

Which function below represents the arithmetic sequence 3, 7, 11, 15...?

Mathematics
2 answers:
mr Goodwill [35]3 years ago
4 0
D) f(n) = 3 + 4(n-1)

3 = 1st term
4 = common difference among the terms
n = term number you are looking for.

To check: 3, 7, 11, 15, ...

f(1) = 3 + 4(1-1) = 3 + 4(0) = 3 + 0 = 3
f(2) = 3 + 4(2-1) = 3 + 4(1) = 3 + 4 = 7
f(3) = 3 + 4(3-1) = 3 + 4(2) = 3 + 8 = 11
f(4) = 3 + 4(4-1) = 3 + 4(3) = 3 + 12 = 15
never [62]3 years ago
4 0

Answer:

its d

Step-by-step explanation:

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Round to the nearest 10th if necessary
Lana71 [14]

Answer:

38.7°.

Step-by-step explanation:

from trigonometrical ratios

SO TO CA

H A H.

Tan°=opposite/adjacent.

Tan°=48/60

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6 0
3 years ago
What is the measure of MBT?
s344n2d4d5 [400]

\\ \sf\longmapsto 17x-6=5x+18

\\ \sf\longmapsto 17x-5x=18+6

\\ \sf\longmapsto 12x=24

\\ \sf\longmapsto x=\dfrac{24}{12}

\\ \sf\longmapsto x=2

Now

\\ \sf\longmapsto \angle{MBT}

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7 0
3 years ago
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Answer:

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Step-by-step explanation:

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4 0
4 years ago
The life in hours of a biomedical device under development in the laboratory is known to be approximately normally distributed.
bearhunter [10]

Answer:

Kindly check explanation

Step-by-step explanation:

H0 : μ = 5500

H1 : μ > 5500

The test statistic assume normal distribution :

Test statistic :

(Xbar - μ) ÷ s/sqrt(n)

(5625.1 - 5500) ÷ 226.1/sqrt(15) = 2.1429 = 2.143

Pvalue from test statistic :

The Pvalue obtained using the calculator at df = 15 - 1 = 14 is 0.025083

α = 0.05

Since ;

Pvalue < α

0.025083 < 0.05 ; Reject H0

The confidence interval :

Xbar ± Tcritical * s/sqrt(n)

Tcritical at 95% = 1.761 ;

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Lower boundary : (5625.1 - 102.805) = 5522.295

(5522.295 ; ∞)

The hypothesized mean does not occur within the constructed confidence boundary. HENCE, there is significant eveidbce to support the claim that the true mean life of a biomedical device is greater than 5500

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3 years ago
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Westkost [7]
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