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2 years ago

If (0.2)x = 2 and log 2 = 0.3010, then the value of x to the nearest tenth is:

Mathematics

2 answers:

spayn [35]2 years ago

8 0

Answer:

(0.2)x = 2.

Taking log on both sides

log (0.2)x = log 2.

x log (0.2) = 0.3010, [since log 2 = 0.3010].

x log (2/10) = 0.3010.

x [log 2 - log 10] = 0.3010.

x [log 2 - 1] = 0.3010,[since log 10=1].

x [0.3010 -1] = 0.3010, [since log 2 = 0.3010].

x[-0.699] = 0.3010.

x = 0.3010/-0.699.

x = -0.4306….

x = -0.4 (nearest tenth)

Answer: (c)

Step-by-step explanation:

hope this helps

Triss [41]2 years ago

5 0

Answer:

Solution:

(0.2)x = 2.

Taking log on both sides

log (0.2)x = log 2.

x log (0.2) = 0.3010, [since log 2 = 0.3010].

x log (2/10) = 0.3010.

x [log 2 - log 10] = 0.3010.

x [log 2 - 1] = 0.3010,[since log 10=1].

x [0.3010 -1] = 0.3010, [since log 2 = 0.3010].

x[-0.699] = 0.3010.

x = 0.3010/-0.699.

x = -0.4306….

x = -0.4 (nearest tenth)

Answer is option (c) -0.4

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a) $\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And replacing we got:

$\bar X = 27.2$

b) For this case we have n =48 observations and we can calculate the median with the average between the 24th and 25th values on the dataset ordered.

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Step-by-step explanation:

For this case we have the following dataset:

23.6, 26.5, 28.6, 28.6, 23.7, 25.3, 24.9, 28.7, 26.7, 32.4, 20.4, 23.9, 32.0, 32.5, 23.7, 26.1, 21.7, 26.1, 38.4, 24.1, 20.5, 25.2, 31, 26.4, 27.1 ,25.1, 25.1, 23.7, 23.9, 32.9, 28.8, 25.6, 28.8, 28.4, 32, 27.6, 29.6, 44.1, 27.1, 24.7, 22.3, 24.3, 23.3, 27.4, 20.4, 25.1, 34.5, 33.1

Part a

We can calculate the mean with the following formula:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And replacing we got:

$\bar X = 27.2$

Part b

For this case we have n =48 observations and we can calculate the median with the average between the 24th and 25th values on the dataset ordered.

For this case the median would be:

$Median = \frac{26.1+26.4}{2}=26.25 \approx 26.3$

Part c

For this case the mode would be:

$Mode= 23.2, 25.1$

And both with a frequency of 3 so then we have a bimodal distribution for this case

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