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masya89 [10]
3 years ago
14

I am not sure on how to do #5

Mathematics
1 answer:
statuscvo [17]3 years ago
5 0
It is too blurry to read please correct me if this is wrong question
a     =17,a      =3 is that what the question says?
  x-1          ,-1

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Whats -5/6 divided by 3/10
enyata [817]

Answer:

-25/9

Step-by-step explanation:

-5/6 : 3/10

-->(cross multibly)

=-5 *10 : 6*3

=-50/ 18

=-25/9

6 0
3 years ago
The driver of a car traveling at 54ft/sec suddenly applies the brakes. The position of the car is s=54t-3t^2, t seconds afyer th
castortr0y [4]
  <span>s = 60t - 3t^2 
v = ds/dt = 60 - 6t 

when it comes to rest, v = 0, 
so t = 10 sec 

now distance travelled = time taken x average velocity 
= 10 x (60 + 0)/2 
= 300 ft 

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6 0
3 years ago
Solve the following system of equations by substitution. <br> 2x - y = 6<br> x + y = 6
m_a_m_a [10]

y=-x+6

2x-(-x+6)=6

3x=12

x=4

4+y=6

y=2

5 0
3 years ago
Read 2 more answers
Which inequality is represented by this graph?
erastovalidia [21]

Answer:the last one I’m not really sure

8 0
3 years ago
Flow meters are installed in urban sewer systems to measure the flows through the pipes. In dry weatherconditions (no rain) the
ziro4ka [17]

Answer:

a) \frac{(8)(30.23)^2}{20.09} \leq \sigma^2 \leq \frac{(8)(30.23)^2}{1.65}

363.90 \leq \sigma^2 \leq 4430.80

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

19.08 \leq \sigma \leq 66.56

b) For this case we are 98% confidence that the true deviation for the population of interest is between 19.08 and 66.56

Step-by-step explanation:

423.6, 487.3, 453.2, 402.9, 483.0, 477.7, 442.3, 418.4, 459.0

Part a

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}}

On this case we need to find the sample standard deviation with the following formula:

s=sqrt{\frac{\sum_{i=1}^8 (x_i -\bar x)^2}{n-1}}&#10;And in order to find the sample mean we just need to use this formula:&#10;[tex]\bar x =\frac{\sum_{i=1}^n x_i}{n}

The sample deviation for this case is s=30.23

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

df=n-1=9-1=8

The Confidence interval is 0.98 or 98%, the value of \alpha=0.02 and \alpha/2 =0.01, and the critical values are:

\chi^2_{\alpha/2}=20.09

\chi^2_{1- \alpha/2}=1.65

And replacing into the formula for the interval we got:

\frac{(8)(30.23)^2}{20.09} \leq \sigma^2 \leq \frac{(8)(30.23)^2}{1.65}

363.90 \leq \sigma^2 \leq 4430.80

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

19.08 \leq \sigma \leq 66.56

Part b

For this case we are 98% confidence that the true deviation for the population of interest is between 19.08 and 66.56

4 0
3 years ago
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