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4 years ago

Determine if the side lengths 6, 11, and 19 form a triangle. If it is a triangle, classify it by its sides.

1 answer:

4 0

It is not triangle because the one side of triangle should be smaller than the other two sides
Let's check
6+11=17
17<19
Or
11+19=31
9<31

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Someone help me pleaseee

krok68 [10]

Answer:

522

Step-by-step explanation:

The assumption is that the proportion of tagged fish caught is the same as the proportion of all fish that were caught and tagged:

(all fish)/(number tagged) = (sample size)/(number in sample tagged)

all fish = (number tagged) · (sample size)/(number in sample tagged)

all fish = 174 · 39/13

all fish = 522

If all the assumptions are reasonable, the number of fish in the lake is estimated at 522.

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4 years ago

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Galina-37 [17]

B is maybe the answer not really sure

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4 years ago

What value of x will make the triangles similar by the SSS similarity theorem?

Sophie [7]

Answer:

1,5

Step-by-step explanation:

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4 years ago

A meat inspector has randomly selected 30 packs of 95% lean beef. The sample resulted in a mean of 96.2% with a sample standard

kirza4 [7]

Answer:

The upper bound of the 99% prediction level is $98.2$

The 95% confidence interval is $9.7383 < \mu < 10.2617$

Step-by-step explanation:

Considering first question

From the question we are told that

The sample size is n = 30

The sample mean is $\= x = 96.2\%$

The standard deviation is $s = 0.8\%$

Generally the degree of freedom is mathematically represented as

$df = n - 1$

=> $df = 30 - 1$

=> $df = 29$

From the question we are told the confidence level is 99% , hence the level of significance is

$\alpha = (100 - 99 ) \%$

=> $\alpha = 0.01$

Generally from the t distribution table the critical value of at a degree of freedom of is

$t_{\alpha , 29} = 2.462$

Generally the 99% prediction level is mathematically represented as

$\= x \pm [(t_{\alpha , df }) * s * (\sqrt{1 + \frac{1}{ n} } )}]$

Generally the upper bound of the 99% prediction level is mathematically represented as

$\= x + [(t_{\alpha , df }) * s * (\sqrt{1 + \frac{1}{ n} } )}]$

=> $96.2 + (2.462 ) * 0.8 * (\sqrt{1 + \frac{1}{ 30} } )}]$

=> $98.2$

Considering second question

Generally the sample is mathematically represented as

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

=> $\= x = \frac{ 9.8 + 10.2 + \cdots +9.6 }{7}$

=> $\= x = 10$

Generally the standard deviation is mathematically represented as

$\sigma = \sqrt{ \frac{ \sum ( x_ i - \= x)}{n-1} }$

=> $\sigma = \sqrt{ \frac{ ( 9.8 -10)^2 + ( 10.2 -10)^2 + \cdots + ( 9.6 -10)^2 }{7-1} }$

=> $\sigma = 0.283$

Generally the degree of freedom is mathematically represented as

$df = n- 1$

=> $df = 7- 1$

=> $df = 6$

From the question we are told the confidence level is 95% , hence the level of significance is

$\alpha = (100 - 95 ) \%$

=> $\alpha = 0.05$

Generally from the t distribution table the critical value of at a degree of freedom of is

$t_{\frac{\alpha }{2} , 6 } = 2.447$

Generally the margin of error is mathematically represented as

$E = t_{\frac{\alpha }{2} , 6 } * \frac{\sigma }{\sqrt{n} }$

=> $E =2.447* \frac{0.283 }{\sqrt{7} }$

=> $E =0.2617$

Generally 95% confidence interval is mathematically represented as

$\= x -E < \mu < \=x +E$

=> $10 -0.2617 < \mu < 10 + 0.2617$

=> $9.7383 < \mu < 10.2617$

7 0

4 years ago

Herberto is constructing the perpendicular bisector of EF¯¯¯¯¯ . He placed the compass point on point E, opened the compass to g

leva [86]

To draw the perpendicular bisector of any line segment, the compass opening should be the kept constant at any value a little more than half the length of the line segment. Then the compass point should be kept on one end and arcs drawn above and below and then keeping the same compass opening the same as mentioned before, the compass point should be placed on the other end and the arcs should be drawn above and below the segment touching the previous arcs.

Thus, out of the given options, the first option best suits the next step Herberto should take.

6 0

4 years ago

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