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Marrrta [24]
3 years ago
13

Urgent!

Mathematics
2 answers:
natima [27]3 years ago
8 0

Answer:

He would have 60$

Step-by-step explanation:

He saves 2$ in the first day and after that everyday from then on is times 3 that number. in ten days you would do 3 times 10 to get 39 then times that by the 2.

Vladimir79 [104]3 years ago
3 0

Answer:

13,122

Step-by-step explanation: 2 x 3 = 6 6 x 3 = 18 18 x 3= 54 54 x 3 = 162 162 x 3 =486 486 x 3 = 1,458 1,458 x 3 = 4,374 4,374 x 3 = 13,122

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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
3 years ago
Since f is parallel to line g, use the diagram to the right right to answer the following question (I need help with problem D a
Snezhnost [94]

Given,

The line f and g are parallel lines.

a)The measure of angle 2 is 117 degree.

By exterior atlernate angle property,

\begin{gathered} \angle2=\angle7 \\ \angle7=117^{\circ} \end{gathered}

The measure of angle 7 is 117 degree.

b)The measure of angle 4 is 68 degree.

By sum of adjacent angle between two parallel lines property,

\begin{gathered} \angle4+\angle6=180^{\circ} \\ \angle6=180^{\circ}-68^{\circ} \\ \angle6=112^{\circ} \end{gathered}

The measure of angle 6 is 112 degree.

c)The measure of angle 5 is 32 degree.

By alternate interior angle property,

\begin{gathered} \angle4=\angle5^{} \\ \angle4=32^{\circ} \end{gathered}

The measure of angle 4 is 32 degree.

d)The measure of angle 7 is 121 degree.

By corresponding angle property,

\begin{gathered} \angle7=\angle3^{} \\ \angle3=121^{\circ} \end{gathered}

The measure of angle 3 is 121 degree.

3 0
1 year ago
Three polynomials are factored below, but some coefficients and constants are missing. All of the missing values of a, b, c, and
mariarad [96]

9514 1404 393

Answer:

  1. (a, b, c, d) = (1, -4, 1, -2)
  2. (a, b, c, d) = (1, -4, 1, 2)
  3. (a, b, c, d) = (1, -4, 2, 6)

Step-by-step explanation:

Once you remove the common factor from the terms, you are looking for factors of the remaining constant term that have a sum equal to the coefficient of the linear term. These factors are the constants in the binomial factors.

__

1. 8 = (-4)(-2)   ⇒   x^2 -6x +8 = (x -4)(x -2)

  (a, b, c, d) = (1, -4, 1, -2)

__

2. -8 = (-4)(+2)   ⇒   3x^3 -6x^2 -24x = 3x(x -4)(x +2)

  (a, b, c, d) = (1, -4, 1, 2)

__

3. -12 = (-4)(3)   ⇒   2x^2 -2x -24 = 2(x -4)(x +3) = (x -4)(2x +6)

   (a, b, c, d) = (1, -4, 2, 6)  or  (2, -8, 1, 3)

3 0
3 years ago
A band director is trying to decide what music to play for the halftime show. To get a sample, the director divides the band int
docker41 [41]

Answer:

random sampling

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Find the centroid of the region bounded by the given curves y=8sin(4x), y=8cos(4x)
Luba_88 [7]
A cosine is just a sine shifted to the left by π/2. A cosine of 4x is shifted to the left by only π/8 because of the factor 4. Sketch them.

The region we're looking for is this sausage-shaped part between the cos and the sin.

The x intercepts are at π/8 for the cosine and π/4 for the sine. The midpoint between them is at (π/8 + π/4)/2 = 3/16π.

The region is point symmetric around the x axis, so the y coordinate of the centroid is 0.

So the centroid is at (3/16π, 0)
5 0
3 years ago
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