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slega [8]
3 years ago
15

A turtle can walk 1/12 of a kilometer in an hour. The turtle is 1/5 o a kilometer away from a pond. At this speed,how long will

it take the turtle to get to the pond?
Mathematics
1 answer:
eimsori [14]3 years ago
7 0

Answer:

It will take 5/12 hours.

Step-by-step explanation:

Given,

The speed of turtle = \frac{1}{12} km per hour,

Distance covered = \frac{1}{5} km,

Since,

Speed =\frac{Distance}{Time}

\implies Time =\frac{Distance}{Speed}

Hence, the speed of turtle = \frac{\text{1}{12}}{\frac{1}{5}}

=\frac{5}{12} hour.

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Match each term with its definition.
Nesterboy [21]

Answer:

that is the answer

mark me as brainliest kid

8 0
3 years ago
Suppose a geyser has a mean time between irruption’s of 75 minutes. If the interval of time between the eruption is normally dis
lesya [120]

Answer:

(a) The probability that a randomly selected Time interval between irruption is longer than 84 minutes is 0.3264.

(b) The probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is 0.0526.

(c) The probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is 0.0222.

(d) The probability decreases because the variability in the sample mean decreases as we increase the sample size

(e) The population mean may be larger than 75 minutes between irruption.

Step-by-step explanation:

We are given that a geyser has a mean time between irruption of 75 minutes. Also, the interval of time between the eruption is normally distributed with a standard deviation of 20 minutes.

(a) Let X = <u><em>the interval of time between the eruption</em></u>

So, X ~ Normal(\mu=75, \sigma^{2} =20)

The z-score probability distribution for the normal distribution is given by;

                            Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean time between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

Now, the probability that a randomly selected Time interval between irruption is longer than 84 minutes is given by = P(X > 84 min)

 

    P(X > 84 min) = P( \frac{X-\mu}{\sigma} > \frac{84-75}{20} ) = P(Z > 0.45) = 1 - P(Z \leq 0.45)

                                                        = 1 - 0.6736 = <u>0.3264</u>

The above probability is calculated by looking at the value of x = 0.45 in the z table which has an area of 0.6736.

(b) Let \bar X = <u><em>sample time intervals between the eruption</em></u>

The z-score probability distribution for the sample mean is given by;

                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean time between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

           n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is given by = P(\bar X > 84 min)

 

    P(\bar X > 84 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{84-75}{\frac{20}{\sqrt{13} } } ) = P(Z > 1.62) = 1 - P(Z \leq 1.62)

                                                        = 1 - 0.9474 = <u>0.0526</u>

The above probability is calculated by looking at the value of x = 1.62 in the z table which has an area of 0.9474.

(c) Let \bar X = <u><em>sample time intervals between the eruption</em></u>

The z-score probability distribution for the sample mean is given by;

                            Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = population mean time between irruption = 75 minutes

           \sigma = standard deviation = 20 minutes

           n = sample of time intervals = 20

Now, the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is given by = P(\bar X > 84 min)

 

    P(\bar X > 84 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{84-75}{\frac{20}{\sqrt{20} } } ) = P(Z > 2.01) = 1 - P(Z \leq 2.01)

                                                        = 1 - 0.9778 = <u>0.0222</u>

The above probability is calculated by looking at the value of x = 2.01 in the z table which has an area of 0.9778.

(d) When increasing the sample size, the probability decreases because the variability in the sample mean decreases as we increase the sample size which we can clearly see in part (b) and (c) of the question.

(e) Since it is clear that the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is very slow(less than 5%0 which means that this is an unusual event. So, we can conclude that the population mean may be larger than 75 minutes between irruption.

8 0
3 years ago
If x and y vary directly and x=25 when y=100 find x when y=24.<br> PLEASE HELP FASTTTTTTT
Veseljchak [2.6K]

Answer:

x = 6

Step-by-step explanation:

Given that x and y vary directly then the equation relating them is

y = kx ← k is the constant of variation

To find k use the condition x = 25 when y = 100 , then

100 = 25k ( divide both sides by 25 )

4 = k

y = 4x ← equation of variation

When y = 24 , then

24 = 4x ( divide both sides by 4 )

6 = x

6 0
3 years ago
HELPPPPPPPPPPPPPPPPPPPPPPPPPPPPP
andrew11 [14]
Ummm not to freak you out but what’s that in the background?
3 0
3 years ago
What is the initial value in the exponential equation below?<br><br> y =<br><br> = 4000(1.05)*
Svet_ta [14]

Answer:

y = 4000 (when x = 0)

Step-by-step explanation:

The initial value of the exponential equation is:

y = 4000\cdot (1.05)^{0}

y = 4000\cdot 1

y = 4000

8 0
3 years ago
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