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Jobisdone [24]
3 years ago
8

Find the equivalent exponential expression

Mathematics
1 answer:
Artemon [7]3 years ago
4 0
[(-5)^3]^2
= (-5)^(3*2)
=(-5)^6

answer
(-5)^6

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Suppose that 11 inches of wire costs 44cents.
maksim [4K]
6 inches of wire can be bought with 24 cents.

44/11 = 4
4 cents per inch
24/4 = 6
4 0
2 years ago
Read 2 more answers
Suppose the dealer incentive per vehicle for honda's acura brand in 2012 is thought to be bell-shaped and symmetrical with a mea
andreev551 [17]

Answer:

From $1600 to $3400.

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 2500

Standard deviation = 300

What interval of dealer incentives would we expect approximately 99.7% of vehicles to fall within?

By the Empirical Rule, 99.7% fall within 3 standard deviations frow the mean. So

From 2500 - 3*300 = 1600 to 2500 + 3*300 = 3400.

8 0
3 years ago
How many bits are required to represent the decimal numbers in the range from 0 to 999 in straight binary code?
ad-work [718]
Note that powers of 2 can be written in binary as

2^0=1_2
2^1=10_2
2^2=100_2

and so on. Observe that n+1 digits are required to represent the n-th power of 2 in binary.

Also observe that

\log_2(2^n)=n\log_22=n

so we need only add 1 to the logarithm to find the number of binary digits needed to represent powers of 2. For any other number (non-power-of-2), we would need to round down the logarithm to the nearest integer, since for example,

2_{10}=10_2\iff\log_2(2^1)=\log_22=1
3_{10}=11_2\iff\log_23=1+(\text{some number between 0 and 1})
4_{10}=100_2\iff\log_24=2

That is, both 2 and 3 require only two binary digits, so we don't care about the decimal part of \log_23. We only need the integer part, \lfloor\log_23\rfloor, then we add 1.

Now, 2^9=512, and 999 falls between these consecutive powers of 2. That means

\log_2999=9+\text{(some number between 0 and 1})

which means 999 requires \lfloor\log_2999\rfloor+1=9+1=10 binary digits.

Your question seems to ask how many binary digits in total you need to represent all of the numbers 0-999. That would depend on how you encode numbers that requires less than 10 digits, like 1. Do you simply write 1_2? Or do you pad this number with 0s to get 10 digits, i.e. 0000000001_2? In the latter case, the answer is obvious; 1000\times10=10^4 total binary digits are needed.

In the latter case, there's a bit more work involved, but really it's just a matter of finding how many number lie between successive powers of 2. For instance, 0 and 1 both require one digit, 2 and 3 require two, while 4-7 require three, while 8-15 require four, and so on.
8 0
3 years ago
A class has 6 boys and 15 girls.What is the ratio of boys and girls?
Sidana [21]
So if boys is 6 and girls is 15 it would be 6/15, but that can be simplified to 2/5.
8 0
3 years ago
a real estate broker sold your house for $72,00. the brokers commission was 6% of the selling pricw. how much would you get afte
worty [1.4K]
$72,00?

I'm assuming you are saying $7200, but if you are saying $72, just replace.

6% = 0.06

7200 / 0.06 = 432

So you would get $432. But trust me, no house is that cheap lol

3 0
3 years ago
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