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DedPeter [7]
3 years ago
10

Help me plz, its due today!!!

Mathematics
1 answer:
d1i1m1o1n [39]3 years ago
8 0
8/125 is your answer
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The number of bees that visit a plant is 600 times the number of years the plant is alive, where t represents the number of year
zavuch27 [327]
C for sure, lock it down
7 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
1. What is the volume, in cubic centimeters, of the sphere shown below? Use 3.14<br> for Pi
prohojiy [21]

Answer:

113.04 cm³

Step-by-step explanation:

diameter = 6cm

so radius = 3cm

Volume of sphere = \frac{4}3}\pi r^3

= 4 * 3.14 * 3 * 3 *3 / 3

=  4 * 28.26

= 113.04 cm³

6 0
3 years ago
Would appreciate any help!! :D
Salsk061 [2.6K]
73+112+90=275
360-275=85
the angle of x is 85 degrees
(pls mark brainliest)
6 0
3 years ago
..................................
yanalaym [24]

Answer:

The answer is a² .

Step-by-step explanation:

You have to substitute x and y into the expression :

let \: x = a \cos(θ) \\ let \: y = a \sin( θ)

{x}^{2}  +  {y}^{2}  =  {(a \cosθ )}^{2}  +  {(a \sinθ) }^{2}

{x}^{2}  +  {y}^{2}  =  {a}^{2}( {cos}^{2}θ) +  {a}^{2} ({sin}^{2}θ)

Next, you have to apply Basic Trigonometric Identity :

{sin}^{2} θ +  {cos}^{2} θ = 1

{x}^{2}  +  {y}^{2}  =  {a}^{2} ( {cos}^{2} θ +  {sin}^{2} θ)

{x}^{2}  +  {y}^{2}  =  {a}^{2} (1)

{x}^{2}  +  {y}^{2}  =  { a}^{2}

8 0
3 years ago
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