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2 years ago

Look at the set of quadrilaterals in the coordinate plane shown below.

1 answer:

3 0

The true statement about the set of quadrilaterals in the coordinate plane is B. Because quadrilateral ABCD can be reflected across the x -axis, then rotated 90° counterclockwise about the origin, and then dilated about the origin by a scale factor of 2 to obtain quadrilateral A'B'C'D' , then quadrilateral ABCD is similar to quadrilateral A'B'C'D'.

<h3>What is a quadrilateral?</h3>

A quadrilateral is a polygon having four sides, four angles, and four vertices.

In this case, because quadrilateral ABCD can be reflected across the x -axis, then rotated 90° counterclockwise about the origin, and then dilated about the origin by a scale factor of 2 to obtain quadrilateral A'B'C'D' , then quadrilateral ABCD is similar to quadrilateral A'B'C'D'.

Learn more about quadrilateral on:

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Police estimate that 25% of drivers drive without their seat belts. If they stop 6 drivers at random, find the probability tha

Furkat [3]

Answer:

17.80% probability that all of them are wearing their seat belts.

Step-by-step explanation:

For each driver stopped, there are only two possible outcomes. Either they are wearing their seatbelts, or they are not. The drivers are chosen at random, which mean that the probability of a driver wearing their seatbelts is independent from other drivers. So we use the normal probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Police estimate that 25% of drivers drive without their seat belts.

This means that 75% wear their seatbelts, so $p = 0.75$

If they stop 6 drivers at random, find the probability that all of them are wearing their seat belts.

This is $P(X = 6)$ .

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 6) = C_{6,6}.(0.75)^{6}.(0.25)^{0} = 0.1780$

17.80% probability that all of them are wearing their seat belts.

3 0

3 years ago

Read 2 more answers

Ethan rollerbladed a total of 623 \text{ km}623 km623, start text, space, k, m, end text over ddd days. He rollerbladed the same

Sever21 [200]

Answer:

Ethan rollerbladed each day $\frac{623}{d}$ kilometers.

Step-by-step explanation:

Given:

Ethan rollerbladed a total of 623 km over d days.

Now, to find the kilometers Ethan rollerblade each day.

Total number of distance rollerbladed = 623 km.

Total number of days = $d.$

<u><em>As, Ethan rollerbladed each day the same distance.</em></u>

Now, to get the distance Ethan rollerbladed in each day we divide total number of distance rollerbladed by total number of days:

$Ethan\ rollerbladed\ each\ day\ =\ \frac{Total\ number\ of\ distance\ rollerbladed}{Total\ number\ of\ days}$

$Ethan\ rollerbladed\ each\ day\ =\ \frac{623}{d}$

Therefore, Ethan rollerbladed each day $\frac{623}{d}$ kilometers.

5 0

3 years ago

B) The cost price of an article was Rs 4000.Find the marked price of the

sweet [91]

Answer:

Market price = Rs. 6,000

Step-by-step explanation:

Given:

Cost price = R.s 4,000

Profit after discount = 20%

Discount = 20%

Find:

Market price

Computation:

Sales price = Cost price[1+Profit after discount]

Sales price = 4,000[1+20%]

Sales price = 4,800

Market price = Sales price[100/(100 - Discount)]

Market price = 4,800[100/(100 - 20%)]

Market price = Rs. 6,000

7 0

3 years ago

∫c<br> x sin y ds, C is the line segment from (0, 1) to (3, 5)

Mnenie [13.5K]

Parameterize the line segment $C$ by

$\mathbf r(t)=\langle x(t),y(t)\rangle=(1-t)\langle0,1\rangle+t\langle3,5\rangle=\langle3t,1+4t\rangle$

where $t\in[0,1]$ . Then

$\mathrm ds=\|\mathbf r'(t)\|\,\mathrm dt$
$\mathrm ds=\|\langle3,4\rangle\|\,\mathrm dt$
$\mathrm ds=5\,\mathrm dt$

So the integral is

$\displaystyle\int_Cx\sin y\,\mathrm ds=5\int_0^1x(t)\sin y(t)\,\mathrm dt$
$=\displaystyle5\int_0^13t\sin(1+4t)\,\mathrm dt$
$=\dfrac{15}{16}(\sin5-\sin-4\cos5)\approx-2.7516$

3 0

3 years ago

HURRY, PLEASE HELP!

Lynna [10]

Answer:

A) y=-3sin(x)

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers