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3 years ago

If an antelope runs 127feet in 10 seconds how fast is the antelope running?

Mathematics

1 answer:

Papessa [141]3 years ago

5 0

Velocity (how fast) = distance over time

127/10 = 12.7 feet per sec

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The height of a free falling object at time t can be found using the function, h(t) = - 12t2 + 36t. Where h(t) is the height in

never [62]

For this case we have the following equation:
h (t) = - 12t2 + 36t
When the object hits the ground we have:
- 12t2 + 36t = 0
We look for the roots of the polynomial:
t1 = 0
t2 = 3
Therefore, the time it takes the object to hit the ground is:
t = 3 s
Answer:
the time when the object hits the ground is:
t = 3 s

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3 years ago

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Which describes the calculations that could be used to solve this problem?

Darya [45]

Answer:

Subtract 37 – 13. Then divide the difference by 4.

Step-by-step explanation:

After the box:

37 - 13 left

No. of bags:

(37 - 13)/4

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3 years ago

Grap f(x)=x^2+1/3(x-8)

gulaghasi [49]

Answer:

2019

Step-by-step explanation:

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2 years ago

Factorize completely, x2 - 25

Furkat [3]

(x+5)(x-5) , You use difference of two squares to do this. Remember that if the sign in the equation was (+) then you wouldn’t be able to continue factoring. You can only factor equations like these if there is a subtraction sign.

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2 years ago

Two part-time instructors are hired by the Department of Statistics and each is assigned at random to teach a single course in p

scZoUnD [109]

Answer:

The probability that they will teach different courses is $\frac{2}{3}$ .

Step-by-step explanation:

Sample space is a set of all possible outcomes of an experiment.

In this case we will write the sample space in the form (x, y).

Here x represents the course taught by the first part-time instructor and y represents the course taught by the second part-time instructor.

Denote every course by their first letter.

The sample space is as follows:

S = {(P, P), (P, I), (P, S), (I, P), (I, I), (I, S), (S, P), (S, I) and (S, S)}

The outcomes where the the instructors will teach different courses are:

s = {(P, I), (P, S), (I, P),(I, S), (S, P) and (S, I)}

The probability of an events E is the ratio of the number of favorable outcomes to the total number of outcomes.

$P(E)=\frac{n(E)}{N}$

Compute the probability that they will teach different courses as follows:

$P(\text{Different courses})=\frac{n(s)}{n(S)}=\frac{6}{9}=\frac{2}{3}$

Thus, the probability that they will teach different courses is $\frac{2}{3}$ .

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2 years ago