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3 years ago

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The price of a cup of coffee has risen to $2.40 today. Yesterday's price was $2.15. Find the percentage increase. Round your ans

wer to the nearest tenth of a percent.
Thank uuu :))))

1 answer:

grigory [225]3 years ago

7 0

Answer:

11.6%

Step-by-step explanation:

(new - old)/(old) = 25/215 = .116 = 11.6%

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In the game of roulette, a wheel consists of 38 slots numbered 0, 00, 1, 2,..., 36. To play the game, a metal ball is spun a

Oxana [17]

Answer:

$E(X)=-0.0526$

$Sd(X)=5.763$

Step-by-step explanation:

In statistics and probability analysis, the expected value "is calculated by multiplying each of the possible outcomes by the likelihood each outcome will occur and then summing all of those values".

The variance of a random variable Var(X) is the expected value of the squared deviation from the mean of X, E(X).

And the standard deviation of a random variable X is just the square root of the variance.

Let X be a random variable which denotes the money you may win or lose on each spin.

In the game of roulette, a wheel consists of 38 slots. To play the game, a metal ball is spun around the wheel and is allowed to fall into one of the numbered slots. If the number of the slot the ball falls into matches the number you selected, we win $35; otherwise we lose $1. So we have just one possibility to win and 37 of lose on an individual game.

___________________________

X P(X)

___________________________

35 1/38

-1 37/38

___________________________

In order to calculate the expected value we can use the following formula:

$E(X)=\sum_{i=1}^n X_i P(X_i)$

And if we use the values obtained we got:

$E(X)=(35)*(\frac{1}{38})+(-1)(\frac{37}{38})=-0.0526$

In order to find the standard deviation we need to find first the second moment, given by :

$E(X^2)=\sum_{i=1}^n X^2_i P(X_i)$

And using the formula we got:

$E(X^2)=(35^2)*(\frac{1}{38})+((-1)^2)(\frac{37}{38})=33.211$

Then we can find the variance with the following formula:

$Var(X)=E(X^2)-[E(X)]^2 =33.211-(-0.0526)^2 =33.208$

And then the standard deviation would be given by:

$Sd(X)=\sqrt{Var(X)}=\sqrt{33.208}=5.763$

4 0

2 years ago

myles and stephen participated in the long jump at the track meet . Myles jumped 4 7/10 meters. Stephens jump was 3 4/5 meters w

Tasya [4]

You would have to subtract both of them and you will get 9/10 stephens jump was further

4 0

2 years ago

Solve absolute value equation 2|x-3|-8=2

Semmy [17]

Answer:

$x=8, -2$

Step-by-step explanation:

$2|x-3|-8=2$

$2|x-3|=10$

$|x-3|=5$ $|x-3|=-5$

$x-3=5$ $x-3=-5$

$x=8$ $x=-2$

7 0

3 years ago

4(1+ 2y) = 12y. Solve for y.

Alja [10]

Answer:

y=1

Step-by-step explanation:

4(1+2y)=12y

4*1 + 4*2y=12y

4 + 8y=12y

4=12y - 8y

4=4y

4/4=y

1=y

8 0

2 years ago

Read 2 more answers

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

2 years ago