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skelet666 [1.2K]
2 years ago
13

Jack had 15 rosebushes he has the same number of yellow,red,and pink bushes,and 3 multicolored bushes find the number of yellow

rosebushes jack has
Mathematics
2 answers:
pogonyaev2 years ago
8 0

Alright, so firstly, we are gonna start with the amount of multicolored bushes Jack has. He currently has 3 multicolored bushes. That's 3 rosebushes off the list. If you read carefully, it says "he has the same number of yellow,red,and pink bushes, and 3 multicolored bushes".

This is simple math.

So, if we do 4+4+4..it would turn out to 12. What's the other number that will go into it to prove this calculation is correct? 3!

This number represents the amount of multicolored bushes there is. The total amount will come out to 15 rosebushes. So, it turns out to 4 because if the rosebushes are all the same number and the answer is 4, there is 4 rosebushes for each color.

So overall,

Jack has 4 yellow rosebushes.

Need any more help? Comment below this answer and tell me!

If you would rather contact me in a different way you may.

Need my email or Discord? Tell me! I'd be glad to help anybody in any way.

- Kana

Masteriza [31]2 years ago
5 0

Answer:

4 yellow rosebushes

Step-by-step explanation:

Givens(What the question tells us):

m = 3(There are 3 multicolored bushes)

r + y + p = 15 - m( total red bushes plus total yellow bushes plus total pink bushes equals 15 - the amount of multicolored bushes)

r = y = p (total red bushes equals total yellow bushes which also equals total pink bushes)

Since we know m =3 we can substitute m for 3 in the second equation:

r + y + p = 15 - 3.

r + y + p = 12

Now since r, y, and p are equal we can divide the total amount left (12) by the 3 kinds of roses left to get our answer:

12/ 3 = 4 red, 4 yellow, and 4 pink

Since we are solving for the amount of yellow the answer is:

4 yellow rosebushes

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(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.

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2 years ago
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