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vovikov84 [41]
3 years ago
7

What is 8 more than twice the sum of a number and 3 is the same as -6?

Mathematics
1 answer:
a_sh-v [17]3 years ago
8 0
This can be solver by creating a small equation with x being the number and solving for x. The equation would be 8 + 2( x + 3 ) = -6
Solve for x:
8 + 2( x + 3 ) = -6
Distribute the 2
8+ (2x+6)=-6
Subtract 8
2x+6= -14
Add 6
2x= 8
Divide by 2
x=4

So the number you are looking for is 4

I hope this helps!
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Svetach [21]

Answer:

I believe the answer is- The mean and MAD can accurately describe the "typical" value in the symmetric data set.

Step-by-step explanation:

The other answers don't make sense because the mean and MAD are being used for symmetrical distributions and asymmetrical means uneven distributions.

6 0
3 years ago
Read 2 more answers
Find a solution to y″+9y=6sin(3t).
QveST [7]
 <span>y" + 9y = 0 
m^2 + 9 = 0 
m = ± 3i 

yH = C1 cos 3t + C2 sin 3t 

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3 0
3 years ago
What is 1.08 to the power of <img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B5%7D" id="TexFormula1" title="\frac{1}{5}" alt="
OlgaM077 [116]

Answer:

about \: 1.01551

Step-by-step explanation:

No it isn't, because:

{1.08}^{ \frac{1}{5} }  =  { \frac{108}{100} }^{ \frac{1}{5} }  =  { \frac{27}{25} }^{ \frac{1}{5} }  =  \sqrt[5]{ \frac{27}{25} }  = 1.01551

6 0
3 years ago
83 random samples were selected from a normally distributed population and were found to have a mean of 32.1 and a standard devi
arlik [135]

Answer:

\frac{(82)(2.4)^2}{104.139} \leq \sigma^2 \leq \frac{(82)(2.4)^2}{62.132}

4.525 \leq \sigma^2 \leq 7.602

Now we just take square root on both sides of the interval and we got:

2.127 \leq \sigma \leq 2.757

Step-by-step explanation:

Information given

\bar X=32.1 represent the sample mean

\mu population mean (variable of interest)

s=2.4 represent the sample standard deviation

n=83 represent the sample size  

Confidence interval

The confidence interval for the population variance is given by the following formula:

\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}

The degrees of freedom given by:

df=n-1=8-1=7

The confidence level is 0.90 or 90%, the value of \alpha=0.1 and \alpha/2 =0.05, and we can use excel, a calculator or a table to find the critical values.  

The excel commands would be: "=CHISQ.INV(0.05,82)" "=CHISQ.INV(0.95,82)". so for this case the critical values are:

\chi^2_{\alpha/2}=104.139

\chi^2_{1- \alpha/2}=62.132

The confidence interval is given by:

\frac{(82)(2.4)^2}{104.139} \leq \sigma^2 \leq \frac{(82)(2.4)^2}{62.132}

4.525 \leq \sigma^2 \leq 7.602

Now we just take square root on both sides of the interval and we got:

2.127 \leq \sigma \leq 2.757

5 0
3 years ago
I'll mark brainliest pls solve it。◕‿◕。<br><img src="https://tex.z-dn.net/?f=y%20%3D%202%28%20-%203%29%20%2B%203" id="TexFormula1
USPshnik [31]

Answer:

Multiply , add and then you get your solution.

Step-by-step explanation:

y=2(-3)+3

y= -6+3

y= -3

5 0
2 years ago
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