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viva [34]
2 years ago
12

How would you solve #2 and #3?

Mathematics
1 answer:
Pani-rosa [81]2 years ago
6 0
Connect the dots to each other
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1. Solve for x.
Allushta [10]

Answer:

i am able to do this only

hope this might help you

3 0
3 years ago
What is the answer to this question I’m stumped.
romanna [79]

Answer:

6x = 16

Step-by-step explanation:

We can write this as a proportion

6         8

---- = ----------

2         x

Using cross products

6x = 2*8

6x = 16

6 0
3 years ago
Read 2 more answers
Round 75.434 to the nearest hundredth
nordsb [41]
Answer:75.43

Explanation:
4 0
2 years ago
Read 2 more answers
((-2^2)(-1^-3))•((-2^-3)(-1^5))^-2
Vitek1552 [10]

Answer:

  256

Step-by-step explanation:

A calculator works well for this.

_____

None of the minus signs are subject to the exponents (because they are not in parentheses, as (-1)^5, for example. Since there are an even number of them in the product, their product is +1 and they can be ignored.

1 to any power is still 1, so the factors (1^n) can be ignored.

After you ignore all of the things that can be ignored, your problem simplifies to ...

  (2^2)(2^-3)^-2

The rules of exponents applicable to this are ...

  (a^b)^c = a^(b·c)

  (a^b)(a^c) = a^(b+c)

Then your product simplifies to ...

  (2^2)(2^((-3)(-2)) = (2^2)(2^6)

  = 2^(2+6)

  = 2^8 = 256

3 0
3 years ago
Assuming that the sample mean carapace length is greater than 3.39 inches, what is the probability that the sample mean carapace
joja [24]

Answer:

The answer is "".

Step-by-step explanation:

Please find the complete question in the attached file.

We select a sample size n from the confidence interval with the mean \muand default \sigma, then the mean take seriously given as the straight line with a z score given by the confidence interval

\mu=3.87\\\\\sigma=2.01\\\\n=110\\\\

Using formula:

z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}  

The probability that perhaps the mean shells length of the sample is over 4.03 pounds is

P(X>4.03)=P(z>\frac{4.03-3.87}{\frac{2.01}{\sqrt{110}}})=P(z>0.8349)

Now, we utilize z to get the likelihood, and we use the Excel function for a more exact distribution

=\textup{NORM.S.DIST(0.8349,TRUE)}\\\\P(z

the required probability: P(z>0.8349)=1-P(z

4 0
2 years ago
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