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andriy [413]
3 years ago
11

horace is running a race. he begins to wonder how fast he is running. which unit rates would be reasonable for horace to use to

describe how fast he is running? select each correct answer. mm/min mi/h in./h km/h
Mathematics
2 answers:
tresset_1 [31]3 years ago
8 0
<span>Horace is running a race. he begins to wonder how fast he is running.
</span>km/h will best describe how fast he is running.

Naily [24]3 years ago
7 0

Answer: The reasonable units that could measure how fast is Horace running, are mi/h and km/h.

Step-by-step explanation:

That is because the velocity is defined as the ratio between the change in the length in an interval of time. We can write this algebraically as:

Velocity (or the same "how fast is Horace running") = ΔL/Δt

So, in that way, Horace could measure his velocity just with a units tan can stablish a big difference of length on the time. For example, it would be easy for him to measure his velocity since the time that he expends while he runs one kilometer. Otherwise, it would be very difficult for him to measure his velocity if he just runs one-millimeter o one-inch, because it will be so difficult too, to measure the time that he expends running that distance

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If a line perpendicular to 2y=x+5 that passes through (2,1) using the slope- intercept form
arlik [135]
Slope of  2y = x + 5:-
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Slope of the line perpendicular to it = -1 / 0.5 = -2

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y- 1 = -2(x - 2)
y = -2x + 4 + 1

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4 0
3 years ago
your 5-year loan of $14 000 is at 4.3% compounded annually. How much will you have paid in total for your loan?
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8 0
3 years ago
(Ross 5.15) If X is a normal random variable with parameters µ " 10 and σ 2 " 36, compute (a) PpX ą 5q (b) Pp4 ă X ă 16q (c) PpX
pochemuha

Answer:

(a) 0.7967

(b) 0.6826

(c) 0.3707

(d) 0.9525

(e) 0.1587

Step-by-step explanation:

The random variable <em>X</em> follows a Normal distribution with mean <em>μ</em> = 10 and  variance <em>σ</em>² = 36.

(a)

Compute the value of P (X > 5) as follows:

P(X>5)=P(\frac{x-\mu}{\sigma}>\frac{5-10}{\sqrt{36}})\\=P(Z>-0.833)\\=P(Z

Thus, the value of P (X > 5) is 0.7967.

(b)

Compute the value of P (4 < X < 16) as follows:

P(4

Thus, the value of P (4 < X < 16) is 0.6826.

(c)

Compute the value of P (X < 8) as follows:

P(X

Thus, the value of P (X < 8) is 0.3707.

(d)

Compute the value of P (X < 20) as follows:

P(X

Thus, the value of P (X < 20) is 0.9525.

(e)

Compute the value of P (X > 16) as follows:

P(X>16)=P(\frac{x-\mu}{\sigma}>\frac{16-10}{\sqrt{36}})\\=P(Z>1)\\=1-P(Z

Thus, the value of P (X > 16) is 0.1587.

**Use a <em>z</em>-table for the probabilities.

8 0
3 years ago
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