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Alex777 [14]
3 years ago
13

What is the value of log

=" {16}^{4} " align="absmiddle" class="latex-formula">
(1 point)​
Mathematics
2 answers:
Dmitriy789 [7]3 years ago
6 0
65,536 I believe is the answer
nignag [31]3 years ago
3 0

9514 1404 393

Answer:

  4·log(16) = 16·log(2) ≈ 4.81648

Step-by-step explanation:

The rules of logarithms are ...

  log(a^b) = b·log(a)

Here, we have ...

  log(16^4) = 4·log(16)

However, we recognize that 16 = 2^4, so this becomes ...

  log(16^4) = 4·(log(2^4)) = 4·4·log(2) = 16·log(2)

The base-10 logarithm of 2 is about 0.30103, so this is ...

  log(16^4) ≈ 16·0.30103 = 4.81648

Choose the form of the answer you need:

  log(16^4) = 4·log(16) = 16·log(2) ≈ 4.81648

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After eating all the leftovers, Chet drove back from Vail to Denver. He
Vladimir79 [104]

Answer:

Driving time Denver to Aspen

Yes, it's about 3-1/2 hours from Denver to Aspen via Glenwood Springs. It's approx. 200 miles, all but the last 40 on I-70 interstate. The driving time will depend upon the weather and what day of the week you're traveling (I-70 can be jammed with skiers on Friday nights and Sat.

Step-by-step explanation:

Im smart

3 0
3 years ago
Most individuals are aware of the fact that the average annual repair cost for an automobile depends on the age of the automobil
o-na [289]

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

Most individuals are aware of the fact that the average annual repair cost for an automobile depends on the age of the automobile. A researcher is interested in finding out whether the variance of the annual repair costs also increases with the age of the automobile. A sample of 26 automobiles 4 years old showed a sample standard deviation for annual repair costs of $120 and a sample of 23 automobiles 2 years old showed a sample standard deviation for annual repair costs of $100. Let 4 year old automobiles be represented by population 1.

State the null and alternative versions of the research hypothesis that the variance in annual repair costs is larger for the older automobiles.

At a 0.01 level of significance, what is your conclusion? What is the p-value?

Answer:

Null hypotheses = H₀ = σ₁² ≤ σ₂²

Alternative hypotheses = Ha = σ₁² > σ₂²

Test statistic = 1.44

p-value = 0.1954

0.1954 > 0.01

Since the p-value is greater than the given significance level therefore, we cannot reject the null hypothesis.

We can conclude that there is no sufficient evidence to support the claim that the variance in annual repair costs is larger for older automobiles.

Step-by-step explanation:

Let σ₁² denotes the variance of 4 years old automobiles

Let σ₂² denotes the variance of 2 years old automobiles

State the null and alternative hypotheses:

The null hypothesis assumes that the variance in annual repair costs is smaller for older automobiles.

Null hypotheses = H₀ = σ₁² ≤ σ₂²

The alternate hypothesis assumes that the variance in annual repair costs is larger for older automobiles.

Alternative hypotheses = Ha = σ₁² > σ₂²

Test statistic:

The test statistic is given by

Test statistic = σ₁²/σ₂²

Test statistic = 120²/100²

Test statistic = 1.44

p-value:

The degree of freedom corresponding to 4 years old automobiles is given by

df₁ = n - 1  

df₁ = 26 - 1  

df₁ = 25

The degree of freedom corresponding to 2 years old automobiles is given by

df₂ = n - 1  

df₂ = 23 - 1  

df₂ = 22

Using Excel to find out the p-value,  

p-value = FDIST(F-value, df₁, df₂)

p-value = FDIST(1.44, 25, 22)

p-value = 0.1954

Conclusion:

When the p-value is less than the significance level then we reject the Null hypotheses

p-value < α   (reject H₀)

But for the given case,

p-value > α

0.1954 > 0.01

Since the p-value is greater than the given significance level therefore, we cannot reject the null hypothesis.

We can conclude that there is no sufficient evidence to support the claim that the variance in annual repair costs is larger for older automobiles.

8 0
3 years ago
One year Ron had the lowest ERA (earned-run average, mean number of runs yielded per nine innings pitched)
Arada [10]

Answer:

Ron's ERA has a z-score of -2.03.

Karla's ERA has a z-score of -1.86.

Due to the lower z-score, Ron had a better yean than Karla relative to their peers.

Step-by-step explanation:

Z-score:

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question:

Since the lower the ERA, the better the pitcher, whoever's ERA has the lower z-score had the better year relative to their peers.

Ron

ERA of 3.06, so X = 3.06

For the males, the mean ERA was 5.086 and the standard deviation was 0.998. This means that \mu = 5.086, \sigma = 0.998

So

Z = \frac{X - \mu}{\sigma}

Z = \frac{3.06 - 5.086}{0.998}

Z = -2.03

Ron's ERA has a z-score of -2.03.

Karla

ERA of 3.28, so X = 3.28

For  the females, the mean ERA was 4.316 and the standard deviation was 0.558. This means that \mu = 4.316, \sigma = 0.558

So

Z = \frac{X - \mu}{\sigma}

Z = \frac{3.28 - 4.316}{0.558}

Z = -1.86

Karla's ERA has a z-score of -1.86.

Which player had the better year relative to their peers, Ron or Karla?

Due to the lower z-score, Ron had a better yean than Karla relative to their peers.

5 0
3 years ago
What is the mean of the numbers?
velikii [3]

Answer:

1.6666666666666...

Step-by-step explanation:

10/6=5/3=1.6666666...

6 0
3 years ago
Determine the measure of ∠FGC. A) 22° B) 70° C) 110° D) 120°
salantis [7]

Answer:

C

Step-by-step explanation:

3 0
3 years ago
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