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SSSSS [86.1K]
3 years ago
10

Help me pleaseeeee somebody ASAP

Mathematics
1 answer:
tatyana61 [14]3 years ago
3 0

Answer:

x = 12

Step-by-step explanation:

In similarity triangles angles are congruent.

∠D = ∠A

x² - 8x = 48

x² - 8x  - 48 = 0

x² + 4x  - 12x - 48 = 0

x(x + 4) - 12(x + 4) = 0

(x +4)(x - 12) = 0

x - 12 = 0     {Ignore x + 4 = 0, as measurements won't have negative values}

x = 12

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Serggg [28]

Number three is a parallelogram

Number four is a rhombus

The attributes of each quadrilateral is that it must have four sides

Number seven is true because if you fold a rectangle in half it will be congruent and rectangles always have at least one parallel side

Number eight is false because a square has all equal length sides and a rectangle does not have all equal sides making the two very different

HOPE THAT HELPS!!!!!!

8 0
3 years ago
Along with this one please
Anarel [89]
X=5.2
Hope this was helpful
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}}
And in order to find the sample mean we just need to use this formula:
[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

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3 years ago
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12345 [234]

Answer:

p=  44 /9

(44 over 9)

Step-by-step explanation:

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3 years ago
The question is in the picture.
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Just download photomath bro, does everything for you.

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