All categories

2 years ago

PLEASE HELP DONT SKIP I NEED YOU

Mathematics

1 answer:

viva [34]2 years ago

6 0

Answer:

72 outfits possible,

1/72 chance of selecting a particular outfit

Step-by-step explanation:

For each of the 3 sweaters, there are 4 shirts to choose from, and for each of those 4 shirts, there are 6 pants to choose from. Therefore, there are totally $3\cdot 4\cdot 6=72$ outfits to choose from (order is fixed and therefore negligible). A specific outfit would represent 1 of these 72 outfits. Therefore, the probability of selecting a particular outfit is $\boxed{\frac{1}{72}}$

You might be interested in

1. What is the sum of 1/9, 2/3, and 5/18? O A. 30 O B. 12/9 O C. 19/18 O D. 4/15

Serhud [2]

Answer:

C. 19/18

Step-by-step explanation:

$\frac{1}{9} + \frac{2}{3} + \frac{5}{18 } = \frac{2}{18} + \frac{12}{18} + \frac{5}{18} = \frac{19}{18}$

So the answer is C. 19/18

6 0

1 year ago

PLEASE HELP ME!! I can’t figure this out

DochEvi [55]

Hello!

The problem has asked that we write a point-slope equation of the line in the image above. Point-Slope Form uses the following formula:

y – $y_{1}$ = m(x – $x_{1}$ )

In this case, M represents the slope while $X_{1}$ and $Y_{1}$ represent the corresponding X and Y values of any given point on the line.

We are given that the slope of the line is - $\frac{5}{7}$ . We also know that any given point on a graph takes the form (x,y). Based on the single point provided in the image above, we can determine that $X_{1}$ is equal to 6 and $Y_{1}$ is equal to 2. Now insert all known values into the point-slope formula above:

y – 2 = - $\frac{5}{7}$ (x – 6)

We have now successfully created an equation based on the information given in the problem above. Looking at the four possible options, we can now come to the conclusion that the answer is C.

I hope this helps!

7 0

3 years ago

How lond did the famouse war of 100 years last ?

tatyana61 [14]

Answer:

It lasted 116 years and saw many major battles – from the battle of Crécy in 1346 to the battle of Agincourt in 1415, which was a major English victory over the French.

6 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

Help me please): I’m stuck like really stuck

julia-pushkina [17]

Area = pi x radius squared... so
a = pi x 10in^2
a = 314.159

6 0

3 years ago