Jane had 35 boxes and she gave to 18 people each how many she has now?

Two standard 6 sided dice are rolled. One is purple and one is green. What is the probability that sum of the two dice on a roll

lana [24]

Answer:

1/6

Step-by-step explanation:

As the purple die has already rolled and the number landed is 6, to find the probability of the sum of the dice being 7, there is only one option for the yellow die:

purple result + yellow result = 7

6 + yellow result = 7

yellow result = 1

So the yellow die needs to roll a 1, over a total of 6 possibilities, so the probability is 1/6

5 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 = 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

3 years ago

Tony separates a number of paintbrushes, x, into 3 equally sized groups. He claims that this is the same as making a group 1/3 t

DochEvi [55]

Answer:

Size of each group = $\frac{x}{3}$

Step-by-step explanation:

Total number of paint brushes = x

Since each the three groups are equally sized, to get the number of paintbrushes in a group, we will have to divide the total number of paint brushes by 3

∴ We have, number of paint brushes in a group = $\frac{Total number of paintbrushes}{3}\\$

= $\frac{x}{3}$

this is also the same thing as saying $\frac{1}{3}$ × $\frac{x}{1}$

This shows that the size of the new group is the same as $\frac{1}{3}$ the size of the original group

4 0

3 years ago

What is the median of the data? 30,29,16,11,16,20,21,12,13,15,15,23,30

tia_tia [17]

Answer:

16

Step-by-step explanation:

ascending order:11,12,13,15,15,16,16,20,21,23,29,30,30

number of terms =13

median=[(n+1)/2] th term

=[(13+1)/2] th term

=[14/2] th tem

=7th term

=16

7 0

3 years ago

during one year of family made five trips to the zoo the distance from their house to the zoo is 46 miles how many miles did the

OlgaM077 [116]

Answer:

460

Step-by-step explanation:

5*46=230 one way but round trip is

230*2=460

4 0

3 years ago