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3 years ago

Rachel jogged along a trail that was 1/4 of a mile long. She jogged along the trail 8 times. How many miles did Rachel jog?

1 answer:

5 0

I am thinking as wisely as I can. I believe that Rachel would have jogged 2 miles, if you think about it. If 1/4 of the trail is jogged each time, and she does that 8 times, 1/4 + 1/4 + 1/4 + 1/4 +1/4 + 1/4 + 1/4 + 1/4= 8/4 or 2. I hope this helped!

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Geometry, please answer question ASAP

Aleonysh [2.5K]

Answer:

Step-by-step explanation:

The inscribed angle IJK is half the measure of its intercepted arc, so

arc IK = 2 × 75° = 150°

The 3 arcs sum to 360° , then

arc IJ = 360° - (150 + 110)° = 360° - 260° = 100° → A

8 0

3 years ago

ABCD is a square of side 3, and E and F are the mid points of sides AB and BC respectively. What is the area of the quadrilatera

Dima020 [189]

Given : - Square ABCD with side 3. E and F as midpoints.
To find : - area of EBFD

Solution : - We have, area of square ABCD = 3 x 3 = 9 units.

Thus, (ar)EBFD = ar ABCD - ar DAE - arDCF

arDAE = 1/2 x base x height

=1/2 x 1.5 x 3 ( AE is 1/2 of AB = 1.5, DA is altitude)
= 2.25

arDFC = 1/2 x base x height
= 1/2 x 1.5 x 3 (FC is 1/2 of BC, DC is altitude)
= 2.25

Thus, (ar) EBFD = arABCD - arDAE - arDCF

= 9 - 2.25 - 2.25

= 4.5 units.

Thus, area of quad EBFD is 4.5 units.

7 0

3 years ago

8-6x=-28 I need help plzzz

Schach [20]

The answer is 6. 8-6(6) = -28.

3 0

3 years ago

Read 2 more answers

You toss a coin and randomly select a number from 0 to 9 what is the probability of getting tails and selecting 3? 0.25 0.95 0 0

andre [41]

The correct answer is .05

To solve, find the decimal of each of the probabilities:

Flipping Tails: 1/2 ----> .5

Picking 3: 1/10-----> .10

Now, multiply them together:

.5 x .10 = .05

Hope this helps!

3 0

4 years ago

Read 2 more answers

A car insurance company has determined that 8% of all drivers were involved in a car accident last year. Among the 15 drivers li

svetlana [45]

Answer:

There is an 11.3% probability of getting 3 or more who were involved in a car accident last year.

Step-by-step explanation:

For each driver surveyed, there are only two possible outcomes. Either they were involved in a car accident last year, or they were not. This means that we solve this problem using binomial probability concepts.

Binomial probability distribution

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

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

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

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

And $\pi$ is the probability of X happening.

In this problem

15 drivers are randomly selected, so $n = 15$ .

A success consists in finding a driver that was involved in an accident. A car insurance company has determined that 8% of all drivers were involved in a car accident last year. This means that $\pi = 0.08$ .

What is the probability of getting 3 or more who were involved in a car accident last year?

This is $P(X \geq 3)$ .

Either less than 3 were involved in a car accident, or 3 or more were. Each one has it's probabilities. The sum of these probabilities is decimal 1. So:

$P(X < 3) + P(X \geq 3) = 1$