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3 years ago

A runner charted how long it took her to run certain distances. Here is the

Mathematics

1 answer:

Serga [27]3 years ago

7 0

Answer:

Step-by-step explanation:8

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Someone please help??

Deffense [45]

Answer:

see below

Step-by-step explanation:

The problem statement seems to presume you have seen an exponential function like this written as ...

f(t) = a0·(1 +r)^t

where a0 is the value corresponding to f(0) and "r" is the fractional rate at which the value increases for each increment of t.

Here, 1+r corresponds to 1.04 in the given function, so r = 0.04 = 4%. When the value is <em>greater than 0</em>, it means there is an <em>increase</em> by that fraction each time t increases by 1.

Here, t is not defined, either, but it would usually be used to represent years in a situation like this. (In other situations, it might represent months, hours, or millenia.)

Hence, the appropriate choice is the one that describes a 4% annual increase.

5 0

3 years ago

An open metal tank of square base has a volume of 123 m^3

snow_lady [41]

Answer:

a) h = 123/x^2

b) S = x^2 +492/x

c) x ≈ 6.27

d) S'' = 6; area is a minimum (Y)

e) Amin ≈ 117.78 m²

Step-by-step explanation:

a) The volume is given by ...

V = Bh

where B is the area of the base, x^2, and h is the height. Filling in the given volume, and solving for the height, we get:

123 = x^2·h

h = 123/x^2

b) The surface area is the sum of the area of the base (x^2) and the lateral area, which is the product of the height and the perimeter of the base.

$S=x^2+Ph=x^2+(4x)\dfrac{123}{x^2}\\\\S=x^2+\dfrac{492}{x}$

c) The derivative of the area with respect to x is ...

$S'=2x-\dfrac{492}{x^2}$

When this is zero, area is at an extreme.

$0=2x -\dfrac{492}{x^2}\\\\0=x^3-246\\\\x=\sqrt[3]{246}\approx 6.26583$

d) The second derivative is ...

$S''=2+\dfrac{2\cdot 492}{x^3}=2+\dfrac{2\cdot 492}{246}=6$

This is positive, so the value of x found represents a minimum of the area function.

e) The minimum area is ...

$S=x^2+\dfrac{2\cdot 246}{x}=(246^{\frac{1}{3}})^2+2\dfrac{246}{246^{\frac{1}{3}}}=3\cdot 246^{\frac{2}{3}}\approx 117.78$

The minimum area of metal used is about 117.78 m².

3 0

2 years ago

An individual repeatedly attempts to pass a driving test. Suppose that the probability of passing the test with each attempt is

vladimir1956 [14]

Answer:

a) Our random variable X="number of tests taken until the individual passes" follows a geomteric distribution with probability of success p=0.25

For this case the probability mass function would be given by:

$P(X= k) = (1-p)^{k-1} p , k = 1,2,3,...$

b) $P(X \leq 3) = P(X=1) +P(X=2) +P(X=3)$

$P(X= 1) = (1-0.25)^{1-1} *0.25 = 0.25$

$P(X= 2) = (1-0.25)^{2-1} *0.25 = 0.1875$

$P(X= 3) = (1-0.25)^{3-1} *0.25 = 0.1406$

And adding the values we got:

$P(X \leq 3) =0.25+0.1875+0.1406=0.578$

c) $P(X \geq 5) = 1-P(X$

And we can find the individual probabilities:

$P(X= 1) = (1-0.25)^{1-1} *0.25 = 0.25$

$P(X= 2) = (1-0.25)^{2-1} *0.25 = 0.1875$

$P(X= 3) = (1-0.25)^{3-1} *0.25 = 0.1406$

$P(X= 4) = (1-0.25)^{4-1} *0.25 = 0.1055$

$P(X \geq 5) = 1-[0.25+0.1875+0.1406+0.1055]= 0.316$

Step-by-step explanation:

Previous concepts

The geometric distribution represents "the number of failures before you get a success in a series of Bernoulli trials. This discrete probability distribution is represented by the probability density function:"

$P(X=x)=(1-p)^{x-1} p$

Let X the random variable that measures the number of trials until the first success, we know that X follows this distribution:

$X\sim Geo (1-p)$

Part a

Our random variable X="number of tests taken until the individual passes" follows a geomteric distribution with probability of success p=0.25

For this case the probability mass function would be given by:

$P(X= k) = (1-p)^{k-1} p , k = 1,2,3,...$

Part b

We want this probability:

$P(X \leq 3) = P(X=1) +P(X=2) +P(X=3)$

We find the individual probabilities like this:

$P(X= 1) = (1-0.25)^{1-1} *0.25 = 0.25$

$P(X= 2) = (1-0.25)^{2-1} *0.25 = 0.1875$

$P(X= 3) = (1-0.25)^{3-1} *0.25 = 0.1406$

And adding the values we got:

$P(X \leq 3) =0.25+0.1875+0.1406=0.578$

Part c

For this case we want this probability:

$P(X \geq 5)$

And we can use the complement rule like this:

$P(X \geq 5) = 1-P(X$

And we can find the individual probabilities:

$P(X= 1) = (1-0.25)^{1-1} *0.25 = 0.25$

$P(X= 2) = (1-0.25)^{2-1} *0.25 = 0.1875$

$P(X= 3) = (1-0.25)^{3-1} *0.25 = 0.1406$

$P(X= 4) = (1-0.25)^{4-1} *0.25 = 0.1055$

$P(X \geq 5) = 1-[0.25+0.1875+0.1406+0.1055]= 0.316$

3 0

3 years ago

Please solve this question

n200080 [17]

Answer:

the answer is b

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

graph PQR with vertices P(-2,3),Q (1,2), and R (3,-1) and its image after the translation ( x,y) -> ( x-2, y - 5)

Furkat [3]

Answer:

(x,y)(x-2,yt)

Step-by-step explanation:

(x,t)(x-2,yt)

we sniled it to the lete by 2 and then we snifted it down by

8 0

2 years ago