Given log_73≈0.5646 and log_716≈1.4248, evaluate the expressions. Please show work using given values.
Answers
Answer:
(a) [tex]\log_72 \approx 0.3562[/tex]
(b) [tex]\log_7(\frac{147}{16} ) = 2[/tex]
Step-by-step explanation:
For the point a:
For this problem we use the information of [tex]\log_716 \approx 1.4248[/tex]
16 can be re-write as [tex]2^4[/tex]. Then you have:
[tex]\log_72^4[/tex]
Using the power property of the logarithm we have:
[tex]\log_ba^c = c\log_ba[/tex]
[tex]\log_72^4 = 4\log_72[/tex]
And this expression is equal to:
[tex]4\log_72 \approx 1.4248\\\log_72\approx \frac{1.4248}{4} \approx 0.3562\\[/tex]
For the point b:
For this problem we use the information of [tex]\log_716 \approx 1.4248[/tex] and [tex]\log_73 \approx 0.5646[/tex]
The first is use the quotient property in the given log.
[tex]\log_c\frac{a}{b} = \log_ca-\log_cb[/tex]
[tex]\log_7\frac{147}{16} = \log_7147 -\log_716[/tex]
We already know the value of [tex]\log_716[/tex] then we have to calculate the value of [tex]\log_7147[/tex].
So we factorize the number 147 in [tex]3 \cdot 7 \cdot 7[/tex] (for factorize the number start with the smaller prime and check if the remainder is 0, it is then is a primer factor of the number).
Now re-write [tex]\log_7147[/tex] as [tex]\log_73 \cdot 7^2[/tex] and by the product property get:
[tex]\log_bmn = \log_b m +\log_b n\\\log_73 \cdot 49 = \log_7 3 +\log_7 49 = 0.5646 + 2 = 2.5646[/tex]
Know replace each value in the subtraction and get the result:
[tex]\log_7147 -\log_716 = 2.5646 - 0.5646 = 2[/tex]
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9514 1404 393
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Step-by-step explanation:
Robert can group his coins into groups consisting of 1 nickel and 4 pennies. The value of each group is 9¢, so the number of groups Robert can make is ...
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Step-by-step explanation:
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Answers
Answers:
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-----------------------------
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-----------------------------
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Answers
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Y= a (x-h)^2 + k
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Answers
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Answers
Answer:
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Step-by-step explanation:
This is a binomial model, and I assume you're meant to use a calculator to solve.
A)
This is what I have in my notes for the binompdf function:
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Using a calculator:
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B)
For this one, use the binomcdf function:
binomcdf(n, p, x) - find the probability of x or fewer successes in n trials given p probabilityIn this case, we're looking for more, not less. However, you can think of the probability of getting more than 6 as just the probability of not getting less than 6.
With a calculator:
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1 - 0.1867 = 0.8133
An 81.33% chance to randomly select at least 6 adults who have little confidence in newspapers.
C)
Same as the above, I don't see a need to explain it all again. Use binomcdf:
binomcdf(10, 0.68, 3) = 0.0155
Notice I used 3 rather than 4. binomcdf calculates x or fewer, or 'rather less than or equal to'. We want just 'less than 4', and that means 'less than or equal to 3'.
That's a 1.55% chance to select less than 4 adults with little confidence in newspapers.