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Anni [7]
3 years ago
5

Changing Bases to Evaluate Logarithms In Exercise, use the change-of-base formula and a calculator to evaluate the logarithm.

Mathematics
1 answer:
Ymorist [56]3 years ago
3 0

Answer:

\frac{3}{2}

Step-by-step explanation:

Changing Bases to Evaluate Logarithms

log_{16}(64)

Apply change of base formula'

log_b(a)= \frac{log a}{log b}

log term should be the numerator and denominator is the log base

log_{16}(64)

log_{16}(a)= \frac{log 64}{log 16}

64 is 4^3  and 16 is 4^2

log_{16}(a)= \frac{log 4^3}{log 4^2}

Move the exponent before log

\frac{3log 4}{2log 4}

top and bottom has same log so cancel it out

\frac{3}{2}

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in a direct variation, y=10when x=5 Write a direct variation equation that shows the relationship between x and y​
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y=2x

Step-by-step explanation:

(10)=2(5)

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2 years ago
If 5 bags cost 255.35 how much would 2 bags cost
rewona [7]

255.35 \div 5 = 51.07

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True.

Step-by-step explanation:

Let's see the definition of in-center of a triangle.

The in-center of a triangle is a point located in the center of the triangle. It is equal distance from all sides of the triangle.

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Is (-4x+9) squared a perfect square
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Step-by-step explanation:

5 0
3 years ago
Suppose that scores on an aptitude test are normally distributed with a mean of 100 and a standard deviation of 4.6. Scores on a
DerKrebs [107]

Answer:

<u>The correct answer is C. Felix's z-score on the aptitude test was 1.30. His z-score on the knowledge test was 2.08. Felix performed better on the knowledge test, comparatively.</u>

Step-by-step explanation:

1. Let's check all the information given to us to answer the question correctly:

Mean of the scores on the aptitude test = 100

Standard deviation of the aptitude test = 4.6

Felix's score on the aptitude test = 106

Mean of the scores on the knowledge test = 70

Standard deviation of the knowledge test = 2.4

Felix's score on the knowledge test = 75

2. Which statement best describes Felix's scores on the two tests comparatively?

Let's recall that z-score in a normal distribution, positive or negative, is the number of times of the standard deviation a certain element is from the mean. If the element is below the mean, then the z-score is negative and if it's above the mean, then the z-score is positive.

Therefore, a score of 106 on the aptitude test, will have the following z-score:

106 - 100 = 6 and it's above the mean.

Now, we calculate the z-score, using the value of the standard deviation this way:

6/4.6 = 1.30

A score of 75 on the knowledge test, will have the following z-score:

75 - 70 = 5 and it's above the mean.

Now, we calculate the z-score, using the value of the standard deviation this way:

5/2.4 = 2.08

The z-scores of Felix were 1.30 on the aptitude test and 2.08 on the knowledge test. He performed better on the aptitude test because 2.08 > 1.30, so the correct statement that best describes Felix's scores on the two tests comparatively is<u> C. Felix's z-score on the aptitude test was 1.30. His z-score on the knowledge test was 2.08. Felix performed better on the knowledge test, comparatively.</u>

6 0
3 years ago
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