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hjlf
2 years ago
10

The table represents an exponential function. A 2-column table has 4 rows. The first column is labeled x with entries 1, 2, 3, 4

. The second column is labeled y with entries three-halves, nine-eigths, StartFraction 27 Over 32 EndFraction, StartFraction 81 Over 128 EndFraction. What is the multiplicative rate of change of the function? Two-thirds Three-fourths Four-thirds Three-halves
Mathematics
2 answers:
Papessa [141]2 years ago
6 0

Answer:

3/4

Step-by-step explanation:

x |  y

1  | 3/2

2 | 9/8

3 | 27/32

4 |  81/128

\frac{\frac{9}{8}}{\frac{3}{2}}=\frac{9}{8} \div \frac{3}{2}=\frac{9}{8} \cdot \frac{2}{3}=\frac{18}{24}=\frac{18 \div 3}{24 \div 3}=\frac{6}{8}=\frac{6 \div 2}{8 \div 2}=\frac{3}{4}

So the multiplicative rate of change of this function is \frac{3}{4} .

givi [52]2 years ago
5 0

Answer:

Option B.

Step-by-step explanation:

The given table is

x           y

1        \frac{3}{2}

2        \frac{9}{8}

3        \frac{27}{32}

4        \frac{81}{128}

We need to find the multiplicative rate of change of the function.

Let multiplicative rate of change is k, then

k=\dfrac{a_2}{a_1}

k=\dfrac{\frac{9}{8}}{\frac{3}{2}}

k=\dfrac{9}{8}\times \dfrac{2}{3}

k=\dfrac{3}{4}

Therefore, the correct option is B.

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<u>ANSWER: </u>

In a data set with a range of 55.4 to 105.4 and 400 observations.86 lies in the 49th percentile.

<u>SOLUTION: </u>

Given, in a data set with a range of 55.4 to 105.4 and 400 observations.

There are 176 observations below the value of 86, and we need to find the percentile for 86.

We know that, percentile formula = \frac{\text {number of observations below the required number}}{\text {total number of observations}} \times 100

Percentile of 86 = \frac{176}{400} \times 100

Since, we cancelled 400 with 100 we get 4 , hence above expression becomes,

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Answer:

a for saturn is 9.55 astronomical units

Step-by-step explanation:

Here, we are interested in calculating the distance from the sun.

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Answer:

<em>The probability that the spark plugs are supposed to have a gap between 3.9mm and 4.3mm.</em>

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Step-by-step explanation:

<u><em>Step(i):-</em></u>

<em>Given that the mean of the Normal distribution = 4.1mm</em>

<em>Given that the standard deviation of the Normal distribution = 0.0075mm</em>

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<u><em>Step(ii):-</em></u>

<em>The probability that the spark plugs are supposed to have a gap between 3.9mm and 4.3mm.</em>

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<u><em>Final answer:-</em></u>

<em>The probability that the spark plugs are supposed to have a gap between 3.9mm and 4.3mm.</em>

<em>P(3.9≤X≤4.3)  = 0.9922</em>

<em></em>

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