1. If △NMK ≅ △TRP, then answer the following questions: (attached is the photo) Solve for X

The sets A,B and C in a universal set consisting of the first 10 lower case letters of the alphabet are: A={a,b,c}, B={b,c,a,e},

Papessa [141]

Answer:

Which item fits in the left side of the table?Criminal versus Civil Cases Criminal cases Civil cases ? . • Accuser is called the plaintiff • Lower burden of proof Defendant may be found liable • Usually heard by a judge only • Accused not guaranteed an attorney

A. No burden of proof

B. Accuser is called the defendant

C. Defendant may be found guilty

D. Usually heard by a jury only

8 0

2 years ago

An airplane flies 2700 miles to Los Angeles to NYC. When the plane flies back to LA, is it flying in a negative direction? use a

dolphi86 [110]

Answer:

Distance = abs(-1700)

Distance = 1700 miles

Step-by-step explanation:

NYC is located East of Los Angels.

The Airplane flies from Los Angeles to the left direction from NYC.

Which is negative direction.

Therefore, distance = -1700

Here the distance cannot be negative even though if it flies in negative direction.

Therefore, we use the absolute value.

The absolute value of a number is always positive.

So, distace = abs( -1700)

Distance = 1700 miles.

Hope this will help you understand the concept.

Thank you.

5 0

3 years ago

In triangle $ABC$, let angle bisectors $BD$ and $CE$ intersect at $I$. The line through $I$ parallel to $BC$ intersects $AB$ and

Umnica [9.8K]

Answer:

41

Step-by-step explanation:

If you work through a series of obscure calculations involving area and the radius of the incircle, they boil down to a simple fact:

... For MN║BC, perimeter ΔAMN = perimeter ΔABC - BC = AB+AC

.. = 17+24 = 41

_____

Wow! Thank you for an interesting question with a not-so-obvious answer.

_____

A little more detail

The point I that you have defined is the incenter—the center of an inscribed circle in the triangle. Its radius is the distance from I to any side, such as BC, for example.

If we use "Δ" to represent the area of the triangle and "s" to represent the semi-perimeter, (AB+BC+AC)/2, then the incircle has radius Δ/s. The area Δ can be computed from Heron's formula by ...

... Δ = √(s(s-a)(s-b)(s-c)) . . . . where a, b, c are the side lengths

For this triangle, the area is Δ = √38480 ≈ 196.1632 units². That turns out to be irrelevant.

The altitude to BC will be 2Δ/(BC), so the altitude of ΔAMN = (2Δ/(BC) -Δ/s). Dividing this by the altitude to BC gives the ratio of the perimeter of ΔAMN to the perimeter of ΔABC, which is 2s.

Putting these ratios and perimeters together, we get ...

... perimeter ΔAMN = (2Δ/(BC) -Δ/s)/(2Δ/(BC)) × 2s

... = (2/(BC) -1/s) × BC × s = 2s -BC

... perimeter ΔAMN = AB +AC

8 0

3 years ago

Evaluate the following numerical expressions. 6 + 23 • 3 =

klasskru [66]

6 + 23(3)

6 + 69

75

Hope this helps! ;)

3 0

3 years ago

Read 2 more answers

An automobile company wants to determine the average amount of time it takes a machine to assemble a car. A sample of 40 times y

aksik [14]

Answer:

A 98% confidence interval for the mean assembly time is [21.34, 26.49] .

Step-by-step explanation:

We are given that a sample of 40 times yielded an average time of 23.92 minutes, with a sample standard deviation of 6.72 minutes.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample average time = 23.92 minutes

s = sample standard deviation = 6.72 minutes

n = sample of times = 40

$\mu$ = population mean assembly time

Here for constructing a 98% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, a 98% confidence interval for the population mean, $\mu$ is;

P(-2.426 < $t_3_9$ < 2.426) = 0.98 {As the critical value of z at 1% level

of significance are -2.426 & 2.426}

P(-2.426 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.426) = 0.98

P( $-2.426 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.426 \times {\frac{s}{\sqrt{n} } }$ ) = 0.98

P( $\bar X-2.426 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.426 \times {\frac{s}{\sqrt{n} } }$ ) = 0.98

98% confidence interval for $\mu$ = [ $\bar X-2.426 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.426 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $23.92-2.426 \times {\frac{6.72}{\sqrt{40} } }$ , $23.92+2.426 \times {\frac{6.72}{\sqrt{40} } }$ ]

= [21.34, 26.49]

Therefore, a 98% confidence interval for the mean assembly time is [21.34, 26.49] .

7 0

3 years ago