Evaluating Logarithms In Exercise, evaluate the logarithm without using a calculator.log4 1/64

Answers

Answer 1
Answer:

Answer:-3

Step-by-step explanation:

Log4 1/64

=log4 4^-3

=-3


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Find the equation of the polynomial function graphed below.A) f(x) = x3 − 3x2 − 6x + 8

B) f(x) = 8x3 − 24x2 − 48x + 64

C) f(x) = x3 + 3x2 − 6x − 8

D) f(x) = 8x3 + 24x2 − 48x − 64

Answers

Answer:

See image

Step-by-step explanation:

Check out the y-intercepts for these equations and look at the y-intercept on the graph. The y-intercept is the point where x=0. So it's mental math you can do in a couple of seconds per equation. If x=0 all the x terms are zero and the y-intercept is just the constant at the end if the equation. I put a note on the graph and the answer choices. The constant is the y-intercept. Only one choice matches. See images.

A) f(x) = x3 − 3x2 − 6x + 8

The probability is 1 in 4,011,000 that a single auto trip in the United States will result in a fatality. Over a lifetime, an average U.S. driver takes 46,000 trips.(a) What is the probability of a

Answers

Answer:

0.0114

Step-by-step explanation:

(a) What is the probability of a fatal accident over a lifetime?

Suppose A be the event of a fatal accident occurring in a single trip.

Given that:

P(1 single auto trip in the United States result in a fatality) = P(A)

Then;

P(A) = 1/4011000

P(A) = 2.493 × 10⁻⁷

Now;

P(1 single auto trip in the United States NOT resulting in a fatality) is:

P(\mathbf{\overline A}) = 1 -  P(A)

P(\mathbf{\overline A}) = 1 - 2.493 × 10⁻⁷

P(\mathbf{\overline A}) = 0.9999997507

However, P(fatal accident over a lifetime) = P(at least 1 fatal accident in lifetime i.e. 46000 trips)

= 1 - P(NO fatal accidents in 46000 trips)

Similarly,

P(No fatal accidents over a lifetime) = P(No fatal accident in the 46000 trips) = P(No fatality on the 1st trip and No fatality on the 2nd trip ... and no fatality on the 45999 trip and no fatality on the 46000 trip)

= [P(\overline A)] ^(46000)  \ \ \  (since \  trips \ are \ independent \ events)

= [0.9999997507]^(46000)

= 0.9885977032

Finally;

P(fatal accident over a lifetime) = 1 -  0.9885977032

P(fatal accident over a lifetime) = 0.0114022968

P(fatal accident over a lifetime) ≅ 0.0114

An architect needs to consider the pitch, or steepness, of a roof in order to ensure precipitation runoff. The graph below showsthe vertical height, y, versus the horizontal distance, x, as measured from the roof peak's support beam.
Roof Steepness
y
14
12
10
8
Vertical Height (feet)
4
2
+X
10 12 14
0
2
4
6
8
Horizontal Distance (feet)
Determine the equation that could be used to represent this situation.

Answers

Answer:

y= -1/2x+10

Step-by-step explanation:

The slope-intercept form of a linear equation is y = mx + b, where m represents the slope and b represents the y-intercept.

For the the given graph, the y-intercept is 10. The slope can be determined by finding the rate of change between any two points on the graph, such as (2,9) and (8,6).

Answer:

The third answer (C).

Step-by-step explanation:

This graph starts at 10. So it needs the +10 at the end.

Also the slope is -1/2 because the graph goes down one, right two. Rise/run.

Please help. Dont put a random answer plz.

Answers

Answer:

Explination:

Work:

Simplify by combining like terms- 9a+3-2a

Answers

Answer: 7a+3

Step-by-step explanation:

9a and 2a are two like terms that we can combine. Because there is a minus sign in front of the 2a, we should subtract 9a from 2a to get

7a. Then because the 3 has plus sign in front of it, we add 7a to 3. BUT because 3 a is NOT a like term, you don’t actually add it, so it looks like this: 7a+3. Hope this helped!

7a+3 because the 9 and 2 are from the same family since they both have a so u combine then and u cant combine 3 with anything so that’s left over

An author argued that more basketball players have birthdates in the months immediately following July​ 31, because that was the age cutoff date for nonschool basketball leagues. Here is a sample of frequency counts of months of birthdates of randomly selected professional basketball players starting with​ January: 390​, 392​, 360​, 318​, 344​, 330​, 322​, 496​, 486​, 486​, 381​, 331 . Using a 0.05 significance​ level, is there sufficient evidence to warrant rejection of the claim that professional basketball players are born in different months with the same​ frequency? Do the sample values appear to support the​ author's claim?

Answers

Answer:

There is sufficient evidence to warrant rejection of the claim that professional basketball players are born in different months with the same​ frequency.

Step-by-step explanation:

In this case we need to test whether there is sufficient evidence to warrant rejection of the claim that professional basketball players are born in different months with the same​ frequency.

A Chi-square test for goodness of fit will be used in this case.

The hypothesis can be defined as:

H₀: The observed frequencies are same as the expected frequencies.

Hₐ: The observed frequencies are not same as the expected frequencies.

The test statistic is given as follows:

 \chi^(2)=\sum{((O-E)^(2))/(E)}

The values are computed in the table.

The test statistic value is \chi^(2)=128.12.

The degrees of freedom of the test is:

n - 1 = 12 - 1 = 11

Compute the p-value of the test as follows:

p-value < 0.00001

*Use a Chi-square table.

p-value < 0.00001 < α = 0.05.

So, the null hypothesis will be rejected at any significance level.

Thus, there is sufficient evidence to warrant rejection of the claim that professional basketball players are born in different months with the same​ frequency.