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m_a_m_a [10]
3 years ago
9

A wheel rotates on an axle 84 times in 6 min. The number of rotations varies directly with the time the wheel rotates.

Mathematics
1 answer:
algol [13]3 years ago
7 0
Do 168/2=84 meaning the time is doubled so its takes 12 mins to rotate 168 times.
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solve the formula for h, and find the height of a trapezoid with an area of 60 m2 and bases of 8m and 12m
Allushta [10]
6, the formula is (b1+b2/2)*h
8+12/2=10
10*h=60 so h=6
3 0
3 years ago
Two tourists went on a hike at dawn. One went from a to b and another one went from b to a. They met at noon but did not stop an
Alecsey [184]
<span>Dawn was at 6 am. Variables a = distance from a to passing point b = distance from b to passing point c = speed of hiker 1 d = speed of hiker 2 x = number of hours prior to noon when dawn is The first hiker travels for x hours to cover distance a, and the 2nd hiker then takes 9 hours to cover that same distance. This can be expressed as a = cx = 9d cx = 9d x = 9d/c The second hiker travels for x hours to cover distance b, and the 1st hiker then takes 4 hours to cover than same distance. Expressed as b = dx = 4c dx = 4c x = 4c/d We now have two expressions for x, set them equal to each other. 9d/c = 4c/d Multiply both sides by d 9d^2/c = 4c Divide both sides by c 9d^2/c^2 = 4 Interesting... Both sides are exact squares. Take the square root of both sides 3d/c = 2 d/c = 2/3 We now know the ratio of the speeds of the two hikers. Let's see what X is now. x = 9d/c = 9*2/3 = 18/3 = 6 x = 4c/d = 4*3/2 = 12/2 = 6 Both expressions for x, claim x to be 6 hours. And 6 hours prior to noon is 6am. We don't know the actual speeds of the two hikers, nor how far they actually walked. But we do know their relative speeds. And that's enough to figure out when dawn was.</span>
8 0
3 years ago
Consider a t distribution with 7 degrees of freedom. Compute P(-1.29 &lt; t &lt; 1.29). Round your answer to at least three deci
marusya05 [52]

Answer:

Step-by-step explanation:

Given that there is a t distribution with 7 degrees of freedom.

P(-1.29 < t < 1.29)=1-0.2380

=0.7620

b) Now there is a different t distribution with 18 df.

When P(t\leq c)=0.05

c=-1.733

7 0
3 years ago
There are eight different jobs in a printer queue. Each job has a distinct tag which is a string of three upper case letters. Th
N76 [4]

Answer:

a. 40320 ways

b. 10080 ways

c. 25200 ways

d. 10080 ways

e. 10080 ways

Step-by-step explanation:

There are 8 different jobs in a printer queue.

a. They can be arranged in the queue in 8! ways.

No. of ways to arrange the 8 jobs = 8!

                                                        = 8*7*6*5*4*3*2*1

No. of ways to arrange the 8 jobs = 40320 ways

b. USU comes immediately before CDP. This means that these two jobs must be one after the other. They can be arranged in 2! ways. Consider both of them as one unit. The remaining 6 together with both these jobs can be arranged in 7! ways. So,

No. of ways to arrange the 8 jobs if USU comes immediately before CDP

= 2! * 7!

= 2*1 * 7*6*5*4*3*2*1

= 10080 ways

c. First consider a gap of 1 space between the two jobs USU and CDP. One case can be that USU comes at the first place and CDP at the third place. The remaining 6 jobs can be arranged in 6! ways. Another case can be when USU comes at the second place and CDP at the fourth. This will go on until CDP is at the last place. So, we will have 5 such cases.

The no. of ways USU and CDP can be arranged with a gap of one space is:

6! * 6 = 4320

Then, with a gap of two spaces, USU can come at the first place and CDP at the fourth.  This will go on until CDP is at the last place and USU at the sixth. So there will be 5 cases. No. of ways the rest of the jobs can be arranged is 6! and the total no. of ways in which USU and CDP can be arranged with a space of two is: 5 * 6! = 3600

Then, with a gap of three spaces, USU will come at the first place and CDP at the fifth. We will have four such cases until CDP comes last. So, total no of ways to arrange the jobs with USU and CDP three spaces apart = 4 * 6!

Then, with a gap of four spaces, USU will come at the first place and CDP at the sixth. We will have three such cases until CDP comes last. So, total no of ways to arrange the jobs with USU and CDP three spaces apart = 3 * 6!

Then, with a gap of five spaces, USU will come at the first place and CDP at the seventh. We will have two such cases until CDP comes last. So, total no of ways to arrange the jobs with USU and CDP three spaces apart = 2 * 6!

Finally, with a gap of 6 spaces, USU at first place and CDP at the last, we can arrange the rest of the jobs in 6! ways.

So, total no. of different ways to arrange the jobs such that USU comes before CDP = 10080 + 6*6! + 5*6! + 4*6! + 3*6! + 2*6! + 1*6!

                    = 10080 + 4320 + 3600 + 2880 + 2160 + 1440 + 720

                    = 25200 ways

d. If QKJ comes last then, the remaining 7 jobs can be arranged in 7! ways. Similarly, if LPW comes last, the remaining 7 jobs can be arranged in 7! ways. so, total no. of different ways in which the eight jobs can be arranged is 7! + 7! = 10080 ways

e. If QKJ comes last then, the remaining 7 jobs can be arranged in 7! ways in the queue. Similarly, if QKJ comes second-to-last then also the jobs can be arranged in the queue in 7! ways. So, total no. of ways to arrange the jobs in the queue is 7! + 7! = 10080 ways

3 0
3 years ago
What is equivalent to log(x)-log(y)
zalisa [80]

Explanation:

log(x) - log(y) = log(x/y)

3 0
3 years ago
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