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2 years ago

10

HELP !!! (will mark brainliest)

2 answers:

Aleksandr [31]2 years ago

5 0

Solution - Verifying Option A:

$(6^{4})^{-5} < (6^{-7}) \times (6^{-3})\\ (6^{-20}) < (6^{-7-3} )\\ 6^{-20}< 6^{-10}$

Result: <u>True (Option A is correct)</u>

Solution - Verifying Option B:

$(6^{4})^{-5} > (6^{-7}) \times (6^{-3})\\ (6^{-20}) > (6^{-7-3} )\\ 6^{-20}> 6^{-10}$

Result: <u>False (Option B is incorrect)</u>

Solution - Verifying Option C:

$(6^{4})^{-5} = (6^{-7}) \times (6^{-3})\\ (6^{-20}) = (6^{-7-3} )\\ 6^{-20}= 6^{-10}$

Result: <u>False (Option C is incorrect)</u>

Option A is correct.

Rasek [7]2 years ago

3 0

Answer:

I think it's A. $(6^{4})^{-5}$

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If 1 is added to the numerator and to the denominator of a fraction,the value obtained is 4/5.If 5 is subtracted from it's numer

Step2247 [10]

Let n and d represent the numerator and denominator of the fraction.
.. (n +1)/(d +1) = 4/5
.. (n -5)/(d -5) = 1/2
Eliminating fractions from these equations gives
.. 5(n +1) = 4(d +1)
.. 2(n -5) = (d -5)
Subtracting the first equation from 4 times the second gives
.. 4(2(n -5)) -(5(n +1)) = 4(d -5) -(4(d +1))
.. 8n -40 -5n -5 = 4d -20 -4d -4
.. 3n = 21
.. n = 7
.. 2(7 -5) = d -5
.. 9 = d

The fraction is 7/9.

4 0

3 years ago

Please help me, I’m brain dead right now:)

Harlamova29_29 [7]

Answer:

A 40%

Step-by-step explanation:

2/5 = 40/100 (or 40%)

So 40 blocks should be filled in :)

Hope this helps!

3 0

3 years ago

Read 2 more answers

Using the simple random sample of weights of women from a data set, we obtain these sample statistics: nequals45 and x overbar

Tomtit [17]

Answer: (141.1, 156.48)

Step-by-step explanation:

Given sample statistics : $n=45$

$\overline{x}=148.79\text{ lb}$

$\sigma=31.37\text{ lb}$

a) We know that the best point estimate of the population mean is the sample mean.

Therefore, the best point estimate of the mean weight of all women = $\mu=148.79\text{ lb}$

b) The confidence interval for the population mean is given by :-

$\mu\ \pm E$ , where E is the margin of error.

Formula for Margin of error :-

$z_{\alpha/2}\times\dfrac{\sigma}{\sqrt{n}}$

Given : Significance level : $\alpha=1-0.90=0.1$

Critical value : $z_{\alpha/2}=z_{0.05}=\pm1.645$

Margin of error : $E=1.645\times\dfrac{31.37}{\sqrt{45}}\approx 7.69$

Now, the 90% confidence interval for the population mean will be :-

$148.79\ \pm\ 7.69 =(148.79-7.69\ ,\ 148.79+7.69)=(141.1,\ 156.48)$

Hence, the 90% confidence interval estimate of the mean weight of all women= (141.1, 156.48)

3 0

3 years ago

Which trend line has been drawn correctly using the divide-center method?

dangina [55]

Answer:

I'm pretty sure its C but some please correct me if I'm wrong.

3 0

3 years ago

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Find MN to the nearest tenth?<br><br> A. 11.34<br> B. 16.1<br> C. 39.6<br> D. 23.8

Nana76 [90]

As the question say this problem is a practice of the law of cosines!

The law of cosine: c^2=a^2+b^2-2ab*cos*c for any side a, b and c

The law can also be written as c=sqrt(a^2+b^2-2abcosc)

Now use this formula!(note cos 100 degrees is about 0.8623)

c=sqrt(15^2+16^2-2*15*16*cos100)

solving this we can MN is about 23.75

Round this to the nearest tenths now!

We get 23.8!

Thus the answer is D

5 0

3 years ago