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3 years ago

What is the answer of 135=-9x+9?

Mathematics

2 answers:

GREYUIT [131]3 years ago

8 0

Answer:

x=-14

Step-by-step explanation:

move all the terms that don't contain <em>x </em>to the right side and solve.

(idk if this is correct but i hope it helps)

Reptile [31]3 years ago

5 0

Answer:

Please mark me brainliest and I hope this helped!

X = -14

Step-by-step explanation:

First, we want to isolate the variable. So we subtract 9 from both sides and that gives us 126= -9. Now we divide both sides by -9. That gives us -14=x. So the answer is:

x = -14

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A group of students estimated the length of one minute without reference to a watch or clock, and the times (seconds) are list

alexandr402 [8]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Yes the students are reasonably good at estimating one minute

Option A is correct

The test statistics is $t = 0.354$

Step-by-step explanation:

From the question we are told that

The set of data is

68, 82 , 38 , 62 , 41, 25 , 57 , 64, 67, 47, 61, 71, 91, 87, 64

The population mean is $\mu = 60$

The level of significance is given as $\alpha = 0.01$

The critical value for this level of significance obtained from the normal distribution table is

$Z_{\alpha } = 2.33$

The null hypothesis is

$H_o : \mu = 60 \ seconds$

The alternative hypothesis is

$Ha : \mu \ne 60 \ seconds$

Generally the sample mean is mathematically represented as

$\= x = \frac{\sum x_i }{n }$

where n = 15

$\= x = \frac{ 68+ 82 + 38 + 62 + 41+ 25 + 57 + 64+67+ 47+ 61+ 71+ 91+ 87+ 64}{15}$

$\= x = 61.67$

The standard deviation is mathematically represented as

$\sigma =\sqrt{ \frac{ \sum (x_i - \= x )^2}{n} }$

substituting values

$\sigma =\sqrt{ \frac{ \sum (68-61.67)^2 + ( 82-61.67 )^2 + (38 -61.67)^2 + 62-61.67)^2 + (41-61.67)^2 +(25-61.67)^2 +( 57+61.67)^2 }{15} }$ $\sqrt{ \frac{ ( \cdot \cdot + ( 64-61.67)^2 + (67-61.67)^2 +(47-61.67)^2 + (61-61.67)^2+ (71-61.67)^2 + (91 -61.67)^2+( 87-61.67)^2 + (64-61.67)^2}{15} }$ => $\sigma = 18.23$

The test statistics is evaluated as

$t = \frac{\= x - \mu }{ \frac{\sigma }{\sqrt{n} } }$

substituting values

$t = \frac{61.67 -60 }{ \frac{18.23 }{\sqrt{ 15} } }$

$t = 0.354$

Now comparing the statistics and the critical value of the level of significance we see that the the test statistics is less than the critical value

Hence the we fail to reject the null hypothesis which mean that these times are from a population with a mean equal to 60 seconds

So we can state that yes the students are reasonably good at estimating one minute given that the sample mean is not far from the population mean

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If f(x)=x^2-2x find f(-3)

Arturiano [62]

Just substitute (-3) wherever you see x
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