What is the ratio of 20 praise and 2000 rupees

Can someone pls pls help meeee

Burka [1]

Answer:

n+7 $\leq$ 9

Step-by-step explanation:

3 0

2 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 = the interval of time between the eruption

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 = 0.3264

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 $\bar X$ = sample time intervals between the eruption

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 = 0.0526

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 $\bar X$ = sample time intervals between the eruption

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 = 0.0222

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

2 years ago

There is a 15% it will rain on both Saturday and Sunday. There is a 60% chance of rain on Saturday. What is the chance of

Artemon [7]

Answer:

25%

Step-by-step explanation:

This question is about conditional probability. Let's say that the probability of raining on Saturday is X=true and the probability of raining on Sunday is Y=true. There is a 15% it will rain on both Saturday and Sunday, to put into the equation it will be:

P(X= true ∩ Y = true) = P(X = true) * P(Y = true)= 0.15

There is a 60% chance of rain on Saturday, mean the equation is

P(X = true) = 0.6

The question is asking for the chance of rain on Sunday or P(Y = true). If we substitute the second equation to first, it will be:

P(X = true) * P(Y = true)= 0.15

0.6* P(Y = true)= 0.15

P(Y = true)= 0.15/0.6

P(Y = true)= 0.25 = 25%

5 0

3 years ago

What is the size of the tv when it’s 66 inches long and 56 inches tall

aleksklad [387]

Area= 3696in squared
Perimeter= 244in

3 0

2 years ago

6. rule (x,y) ---> is the image similar to the original pre-image?

Artyom0805 [142]

Here, we want to check for the relationship between the image and its pre-image

The pre-image is (x,y)

The image is (3x,3y+5)

As we can see, the pre-image is not similar

This is because the transformation applied to the two values are not same

Thus, we have that;

No, the image is not similar to the pre-image as the translations applied to both coordinates are not same

The pre-image was transformed by dilating the x-coordinate of the pre-image by 3 units while the y-coordinate was transformed by dilating the y-coordinate of the pre-image by 3 and translating it upward by 5 units

3 0

1 year ago

What is the ratio of 20 praise and 2000 rupees​

What is the ratio of 20 praise and 2000 rupees