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3 years ago

Select the quadrilateral for which the diagonal is a line of symmetry.

Mathematics

2 answers:

Alika [10]3 years ago

6 0

Answer:

Step-by-step explanation:

Dafna11 [192]3 years ago

3 0

Answer:

Step-by-step explanation:

cut it in half, boom equal

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How many years will it take James to recover his investment?

My name is Ann [436]

I don’t think anyone knows this one not even James

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3 years ago

At a specific point on a highway, vehicles arrive according to a Poisson process. Vehicles are counted in 12 second intervals, a

morpeh [17]

Answer: a) 4.6798, and b) 19.8%.

Step-by-step explanation:

Since we have given that

P(n) = $\dfrac{15}{120}=0.125$

As we know the poisson process, we get that

$P(n)=\dfrac{(\lambda t)^n\times e^{-\lambda t}}{n!}\\\\P(n=0)=0.125=\dfrac{(\lambda \times 14)^0\times e^{-14\lambda}}{0!}\\\\0.125=e^{-14\lambda}\\\\\ln 0.125=-14\lambda\\\\-2.079=-14\lambda\\\\\lambda=\dfrac{2.079}{14}\\\\0.1485=\lambda$

So, for exactly one car would be

P(n=1) is given by

$=\dfrac{(0.1485\times 14)^1\times e^{-0.1485\times 14}}{1!}\\\\=0.2599$

Hence, our required probability is 0.2599.

a. Approximate the number of these intervals in which exactly one car arrives

Number of these intervals in which exactly one car arrives is given by

$0.2599\times 18=4.6798$

We will find the traffic flow q such that

$P(0)=e^{\frac{-qt}{3600}}\\\\0.125=e^{\frac{-18q}{3600}}\\\\0.125=e^{-0.005q}\\\\\ln 0.125=-0.005q\\\\-2.079=-0.005q\\\\q=\dfrac{-2.079}{-0.005}=415.88\ veh/hr$

b. Estimate the percentage of time headways that will be 14 seconds or greater.

so, it becomes,

$P(h\geq 14)=e^{\frac{-qt}{3600}}\\\\P(h\geq 14)=e^{\frac{-415.88\times 14}{3600}}\\\\P(h\geq 14)=0.198\\\\P(h\geq 14)=19.8\%$

Hence, a) 4.6798, and b) 19.8%.

7 0

4 years ago

Which triangle’s unknown side length measures 7 units?

kow [346]

A (A right triangle with a side length of 5 and hypotenuse with length \sqrt{74})

8 0

3 years ago

Read 2 more answers

Describe the rotation of J

Serga [27]

Answer:

The coordinates of J' when rotating by 90° counterclockwise will be: J'(3, 1)

The coordinates of J' when rotating by 90° clockwise will be: J'(-3, -1)

Step-by-step explanation:

Square JKLM with vertices

J(1, -3)

K(5, 0)

L(8, -4)

M(4, -7)

We have to determine the answer for the image J' of the point (1, -3) when we rotate the point by 90° counterclockwise, we need to switch x and y, make y negative.

In other words, the rule to rotate a point by 90° counterclockwise.

P(x, y) → P'(-y, x)

As we are given that J(1, -3), so the coordinates of J' will be:

J(1, -3) → J'(3, 1)

Therefore, the coordinates of J' when rotating by 90° counterclockwise will be: J'(3, 1).

When the point is rotated by 90° clockwise, we need to switch x and y, make x negative.

In other words, the rule to rotate a point by 90° clockwise.

P(x, y) → P'(y, -x)

As we are given that J(1, -3), so the coordinates of J' will be:

J(1, -3) → J'(-3, -1)

Therefore, the coordinates of J' when rotating by 90° clockwise will be: J'(-3, -1)

3 0

3 years ago

Describe how to find the sums -4+2 and -4+(-2) on a number line.

kifflom [539]

For -4+2, plot -4, then move to the right 2 spaces. You should land on -2.

For -4+(-2), plot -4 then move the left 2 spaces. You should land on -6.

3 0

3 years ago

Read 2 more answers