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Thepotemich [5.8K]
3 years ago
11

Using Properties of logarithms In Exercise,Use the properties of logarithms and a fact that In 2≈ 0.6931 and In 3 ≈1.0986 to app

roximate the logarithm.Then use a calculator to confirm your approximation.
(a) In 0.25
(b) In 24
(c) In (12)1/3
(d) In 1/72
Mathematics
1 answer:
Ber [7]3 years ago
6 0

Answer:

Step-by-step explanation:

Given that ln 2 = 0.6931 and ln 3 = 1.0986

We have to find the values for the given numbers

using properties of log

Properties of log are

log a+log b = log ab:  log a-log b = log a/b

a) ln 0.25 = ln 1-ln 4 = -2ln 2 = -1.3862

b) ln 24 = ln 3 (2^3) = 3 ln 2 + ln 3 = 2.0793+1.0986

=3.1779

c) ln (12)^(1/3) = 1/3 ln (2^2*3)  =1/3 (1.3862+1.0986)

= 0.8283

d) ln (1/72) = -ln 72 = -ln 9 -ln 8

= -3 ln 3 - 3ln 2

= -5.2752

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Please help<br><br> Combine like terms.<br> 9y + 5y - 3 = [? ]y + [ ]
almond37 [142]

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3 years ago
5.916 in word and expanded form
zzz [600]
Five. nine thousand and sixteen
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3 years ago
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If a person walks 2.4 miles in 1.5 hours how far can you walk in 6 hours
mestny [16]

Answer: 9.6

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So a person can walk 9.6 miles in 6 hours.

Hope this helps!

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3 years ago
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Last week’s and this week’s low temperatures are shown in the table below. Low Temperatures for 5 Days This Week and Last Week L
STatiana [176]

Question:

Temperatures for 5 Days This Week and Last Week

Low Temperatures This Week (Degrees Fahrenheit) 4 10 6 9 6 Low Temperatures Last Week (Degrees Fahrenheit) 13 9 5 8 5 Which measures of center or variability are greater than 5 degrees? Select three choices.

a) the mean of this week’s temperatures

b) the mean of last week’s temperatures

c) the range of this week’s temperatures

d)the mean absolute deviation of this week’s temperatures

e) the mean absolute deviation of last week’s temperatures

Answer:

a) the mean of this week’s temperatures

b) the mean of last week’s temperatures

c) the range of this week’s temperatures

Step-by-step explanation:

I would be verifying the options a, b, and c in my answer above through calculations which are shown below.

We were given the following data:

Low Temperatures This Week (Degrees Fahrenheit)

4, 10, 6, 9, 6

Low Temperatures Last Week (Degrees Fahrenheit)

13, 9, 5, 8, 5

We are to find which measures of center or variability are greater than 5 degrees.

Option a

The mean of this week’s temperatures

(4+ 10+ 6+9+ 6) °F ÷ 5 = 35 °F ÷5 = 7°F

Option a is correct because it measures of center or variability which is 7 °F is higher than 5°F

Option b

The mean of last week’s temperatures

(13+ 9 + 5 + 8 + 5) °F = 40°F ÷ 5 = 8°F

Option b is correct , because its measure of variability which is 8°F is greater than 5°F.

Option c

the range of this week’s temperatures

This week's temperature is given as

(4, 10, 6, 9, 6) °F

Range is defined as the difference between the highest number and the lowest number

Range of this week's temperature = (10 - 4) °F = 6°F

Hence Option c is correct because it measures of center or variability which is 6°F is greater than 5°F

From the above calculations we can accurately confirm that options a, b, and c are correct because the measures of their center or variability is greater than 5°F

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3 years ago
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