a cheeta runs a speed of 65mi/hr, how many meters/second is that?​

The total distance covered by Mary while walking around the track is 2 1/2 miles.

On a track, Mary and Jerry are working out.

Mary is moving along at a 3 mph pace.

After Mary has been walking for fifteen minutes, Jerry begins to jog at a speed of four miles per hour.

Mary continues to walk as Jerry jogs for two kilometers before they both stop at the same time.

Enter Mary's total walking mileage around the track in miles.

Find the distance M had walked when J starts jogging.

15 min = 0.25 hrs

3 × .25 = .75 miles

Find how long it took J to jog 2 miles at 4 mph

2/4 = .5 hrs

Find how far M can walk in .5 hrs at 3 mph

.5 × 3 = 1.5 miles

Enter the total distance, in miles, that Mary walks around the track

.75 + 1.5 = 2.25 mi or 2 1/4 miles.

Hence distance covered in miles is 2.25 miles.

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Answer:

no

Step-by-step explanation:

she did run/rise instead of rise/run

Answer:

V = 3 * 3 * 7 + 2 * 3 * 5 = 63 + 30 = 93       just adding 2 segments

Answer:

3686

Step-by-step explanation:

Step-by-step explanation:

Brian takes = 43 minutes

Colin takes = 1 hour = 60 minutes

In ratio,

43/60

Answer:

4900 Litres

Step-by-step explanation:

First we need a common demoninator:

3/7=15/35

Then we subtract the two to figure out how much 420 Litres is:

15/35-12/35=3/35

3/35=420 litres

Divide amount by the numerater for thow much 1/x is.

1/35=140

and multiply by the denomenator to get a full number on your fraction, and therefore a full tank.

140 x 35 = 4900

Answer

one line is points (-9,4 ) (1,-2) and both cross over each at (3,4)

Step-by-step explanation:

Honors algebra was a tragic year

True. Time series data refers to a collection of observations gathered over time, where the time dimension is a critical component of the data.

Each data point is linked to a specific point in time. In contrast, cross-sectional data is collected at a single point in time and does not have a time dimension. Therefore, the timing of each observation is crucial in time series data but not as important in cross-sectional data.

Time series data is a sequence of data points indexed in time order. It is used to track change over time.

Cross-sectional data is a snapshot of data at a specific point in time. It is used to compare different groups or variables .

Think about how the time at which each observation is made affects the analysis of time series data and cross-sectional data.

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The areas that needs correction has been attended to and they include the following:

4.)533.8 meters

5.)22686.5 m²

8.)19.06 m

To determine the radius of a circle, the diameter should be divided into two.

For the wheel, the radius is calculated as follows;

Diameter = 150/2

= 85 meters.

For 4.)

The circumference of the wheel with the given radius;

Formula = 2πr

= 2×3.14×85

= 533.8 meters

For 5.)

Area of the wheel = πr²

= 3.14×85×85 = 22686.5 m²

For 8.) The arc length between the two cars;

= circumference/number of compartment

= 533.8/28

= 19.06 m

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Answer:

10.25641...% or 10.3%

Step-by-step explanation:

(86/78)/78 = 8/78

8/78 * 100 = 400/39 = 10.25641...%

Answer:

10

Step-by-step explanation:

2+2=4

4+6= 10

Answer:

The answer is uhhhhhhhh 10

Step-by-step explanation:

The Practice Operations with radical expressions is simplified using the basic of the Algebra:

Algebraic expressions incorporating radicals are known as radical expressions. The root of an algebraic expression makes up the radical expressions (number, variables, or combination of both). The root might be an nth root, a square root, or a cube root. Radical expressions can be made simpler by taking them down to their most basic form and, if feasible, getting rid of all of the radicals.

Radical expressions are simplified by taking them down to their most basic form and, if feasible, altogether deleting the radical. An algebraic expression's numerator and denominator are multiplied by the appropriate radical expression if the denominator contains a radical expression.

Practice Operations with radical expressions are:

1) \(\sqrt{540}\) = \(\sqrt{36 * 15 }\)

= 6√15

2) \(\sqrt[3]{432}\) = \(\sqrt[3]{216*2}\)

= \(6\sqrt[3]{2}\)

3) \(\sqrt[3]{128}\) = \(\sqrt[3]{64*2}\)

= \(4\sqrt[3]{2}\)

4) \(-\sqrt[4]{405}\) = \(-\sqrt[4]{81*5}\)

= \(-3\sqrt[4]{5}\)

5) \(\sqrt[3]{-5000}\) = \(-\sqrt[3]{1000*5}\)

= \(-10\sqrt[3]{5}\)

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The expected number of boxes to be produced is given as 3,000 boxes. So, the correct answer is 0.3, indicating that the takt time in minutes is 0.3 minutes.

The production line operates for two eight-hour shifts each day, which means there are 16 hours of production time available. Since there are 60 minutes in an hour, the total available time in minutes would be 16 hours multiplied by 60 minutes, which equals 960 minutes.

The expected number of boxes to be produced is given as 3,000 boxes.

To calculate the takt time in minutes, we divide the total available time (960 minutes) by the expected number of boxes (3,000 boxes):

\(Takt time = Total available time / Expected number of boxes\)

\(Takt time = 960 / 3,000\)

By performing the calculation, we find that the takt time is approximately 0.32 minutes, which is equivalent to 0.3 minutes rounded to one decimal place.

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The probability that headway is within 1 standard deviation of the mean value is 0.890.

What is mean value?

By dividing the sum of the given numbers by the entire number of numbers, the mean—the average of the given numbers—is determined.

The mean of a discrete probability distribution of a random variable X is equal to the sum over all possible values weighted by the likelihood of each value. To calculate the mean, one must multiply each potential value of X by its probability P(x), then add all of these products.

Given mean value is P( 0.995 ≤ x ≤ 1.445) = F(1.445) - F(0.995).

The cdf for the distribution is

\(F(x)=\left \{ {{1 -\frac{1}{x^6} ,x > 1} \atop {0,x\le 1}} \right.\)

Now calculate the value of F(1.445) and F(0.995).

F(1.445) = 1 - (1/(1.445)⁶)

F(0.995) = 0

Now putting the value of F(1.445) and F(0.995) in  P( 0.995 ≤ x ≤ 1.445) = F(1.445) - F(0.995):

P( 0.995 ≤ x ≤ 1.445)

=1 - (1/(1.445)⁶)  - 0

=0.890

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Answer: that the people he is addressing appreciate intellectualism

Step-by-step explanation: found it on quizlet

Answer:

That the people he is addressing have researched philosophy

Step-by-step explanation:

i said so

Answer:8x=10y

Step-by-step explanation:

A proportional relationship between the number of stickers Xander buys, x, and the total cost, y is 8x=10y.

Given that, Xander pays $8 for 10 skateboard stickers.

Proportional relationships are relationships between two variables where their ratios are equivalent. Another way to think about them is that, in a proportional relationship, one variable is always a constant value times the other. That constant is know as the "constant of proportionality".

Given, the number of stickers Xander buys, x, and the total cost, y.

Now, 8x=10y

⇒ x/y = 10/8

Therefore, a proportional relationship between the number of stickers Xander buys, x, and the total cost, y is 8x=10y.

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Answer:

B)XZ

Used a calculator and just took this quiz

Answer:

B

Step-by-step explanation:

m < y = 180 degrees - 60 degrees - 43 degrees = 77 degrees

77 > 60 > 43

xz >  yz  > xy

XZ is the longest side. So the answer is B

Answer:

its abcd

this is the answer because its asking you for the pattern

hope i helped, TRUST ME ITS RIGHT

The range of the given table would be set of all possible y-values as {-6, -2, 5}​ which is the correct answer would be an option (D).

The domain of the function includes all possible x values of a function, and the range includes all possible y values of the function.

We have been given that table as:

x     y

-3    -6

1     -2

7     5

As per the given table, the required solution would be as:

The range of a function is the set of all possible output values of the function. In this case, the output values of the function are the y-values in the table. The range of the function table is the set of all possible y-values, which in this case is {-6, -2, 5}.

Therefore, the range of the function table is therefore -6 to 5.

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The probability that an arriving student will find at least one other student waiting in line is approximately 0.167.

To solve this problem, we'll use the M/M/1 queueing model with Poisson arrivals and exponential service times. Let's calculate the required values: (a) Percentage of time Judy is idle: The utilization of the system (ρ) is the ratio of the average service time to the average interarrival time. In this case, the average service time is 10 minutes, and the average interarrival time is 12 minutes. Utilization (ρ) = Average service time / Average interarrival time = 10 / 12 = 5/6 ≈ 0.8333

The percentage of time Judy is idle is given by (1 - ρ) multiplied by 100: Idle percentage = (1 - 0.8333) * 100 ≈ 16.67%. Therefore, Judy is idle approximately 16.67% of the time. (b) Average waiting time for a student:

The average waiting time in a queue (Wq) can be calculated using Little's Law: Wq = Lq / λ, where Lq is the average number of customers in the queue and λ is the arrival rate. In this case, λ (arrival rate) = 1 customer per 12 minutes, and Lq can be calculated using the queuing formula: Lq = ρ^2 / (1 - ρ). Plugging in the values: Lq = (5/6)^2 / (1 - 5/6) = 25/6 ≈ 4.17 customers Wq = Lq / λ = 4.17 / (1/12) = 50 minutes. Therefore, on average, a student spends approximately 50 minutes waiting in line.

(c) Average length of the line: The average number of customers in the system (L) can be calculated using Little's Law: L = λ * W, where W is the average time a customer spends in the system. In this case, λ (arrival rate) = 1 customer per 12 minutes, and W can be calculated as W = Wq + 1/μ, where μ is the service rate (1/10 customers per minute). Plugging in the values: W = 50 + 1/ (1/10) = 50 + 10 = 60 minutes. L = λ * W = (1/12) * 60 = 5 customers. Therefore, on average, the line consists of approximately 5 customers.

(d) Probability of finding at least one student waiting in line: The probability that an arriving student finds at least one other student waiting in line is equal to the probability that the system is not empty. The probability that the system is not empty (P0) can be calculated using the formula: P0 = 1 - ρ, where ρ is the utilization. Plugging in the values:

P0 = 1 - 0.8333 ≈ 0.1667. Therefore, the probability that an arriving student will find at least one other student waiting in line is approximately 0.167.

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The set of integers for which p(k) is true is all even integers for which p(k) is proved true if we also prove that p(k)^p(k 2)=> p(k-2).

Considering that p(- 2) and p(0) are valid and expecting that p(k)^p(k+2) => p(k-2), we can decide the arrangement of numbers k for which p(k) is valid by acceptance. In particular, we can show that p(k) is valid for all even numbers k by enlistment, as follows.

To begin with, since p(- 2) and p(0) are valid, the base instance of k = - 2 is valid.

Then, expect that p(k) is valid for some even number k. Then, at that point, utilizing the given ramifications, we have p(k+2) is valid.

At long last, utilizing a similar ramifications once more, we have that p(k-2) is valid. Accordingly, by acceptance, we have shown that p(k) is valid for all even numbers k.

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The p-value is 2 * 0.0708 = 0.1416.

To find the p-value associated with a z test statistic of 1.47 in a two-tailed test, we need to calculate the probability of observing a test statistic as extreme or more extreme than 1.47 in both tails of the standard normal distribution.

The p-value is the probability of observing a test statistic as extreme or more extreme than the observed value. In a two-tailed test, we need to consider both tails.

To find the p-value, we can use a standard normal distribution table or a statistical calculator. In this case, using a standard normal distribution table, we find that the area to the right of 1.47 is 0.0708.

Since it is a two-tailed test, we need to consider both tails. Therefore, the total p-value is twice the area in one tail.

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This form of the equation is useful because it shows that any periodic function of time can be represented as a sum of sine and cosine functions with different amplitudes and phases.

We can show that acos(wt) + bsin(wt) can be written in the form Rsin(wt + φ) where R and φ are constants and R is the amplitude of the wave.

Using the trigonometric identity sin(a+b) = sin(a)cos(b) + cos(a)sin(b), we can write:

acos(wt) + bsin(wt) = R(sin(wt+φ)),

where R = √(a^2+b^2) and φ = tan⁻¹(b/a).

To see how this works, we can use the values of a = 3 and b = 4 as an example:

acos(wt) + bsin(wt) = 3cos(wt) + 4sin(wt)

R = √(a^2+b^2) = √(3^2 + 4^2) = 5

φ = tan⁻¹(b/a) = tan⁻¹(4/3)

So we can rewrite the equation as:

acos(wt) + bsin(wt) = 5sin(wt + tan⁻¹(4/3))

This form of the equation is useful because it shows that any periodic function of time can be represented as a sum of sine and cosine functions with different amplitudes and phases. It is also helpful in analyzing the properties of waves, such as their amplitude and frequency.

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The values of the variables are x = -16.8 and x = 8

Algebraic expressions are defined as expressions that are composed of varaibles, coefficients, terms, constants and factors.

These algebraic expressions are also made up of mathematical operations, such as;

From the information given, we have that;

5x+64=-20

collect the like terms, we get;

5x = -20 - 64

subtract the values

5x = -84

Divide by the coefficient, we have;

x = - 16. 8

Also,

9  +2x=25

collect like terms

2x = 16

Divide by the coefficient

x = 8

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Answer:

6.493×10^ -6

.............exponent= -6

Answer:

16.65 feet

Step-by-step explanation:

Okay, we have three boards that are each 6.85 feet long. If we cut 3.4 feet off one board, that board has:

6.85

−

3.4

=

3.45

feet left

If we cut off 0.5 feet off another board, that board has:

6.85

−

0.5

=

6.35

feet left

So, if we take our three new boards, the total length of plywood left is:

6.85

+

3.45

+

6.35

=

16.65

feet

Hope this helps!