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3 years ago

The ages of the people in attendance at the showing of an award-winning foreign film are shown below.

Mathematics

2 answers:

guapka [62]3 years ago

5 0

Definition: A histogram visualises the distribution of data over a continuous interval or certain time period. Each bar in a histogram represents the tabulated frequency at each interval.

According to the given histogram,

0-20 years occur 15 times,

21-40 years occur 40 times,

41-60 years occur 40 times,

61-80 years occur 15 times.

This shows that the data is symmetrical and the majority of the attendees were between the ages of 21 and 60 (40 times).

Answer: correct choice is B.

Monica [59]3 years ago

3 0

B. The data is symmetrical, showing that the majority of the attendees were between the ages of 21 and 60.

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A. 25.7°

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the sum of straight angle is 180°.

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2 years ago

There was a bag of counters. 45 were large! For every large counters 3/5 were small! The green was 300% than yellow, there were

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2 years ago

Consider the function f ( x ) = 1 − 4 x 2 on the interval [ − 1 , 3 ] . Find the average or mean slope of the function on this i

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Answer:

The average slope is 8.

Step-by-step explanation:

Let f be defined on the closed interval [a, b]. The average slope of f between a and b is the quotient

average slope = $\frac{f(b)-f(a)}{b-a}$

To find the average slope of the function $f(x)=1-4x^2$ on the interval [-1, 3] you have to evaluate your function in the interval and then divide by the interval.

So,

$f(-1)=1-4(-1)^2=-3\\\\f(3)=1-4(3)^2=-35$

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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.

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2 years ago

MATH 15 POINTS, HELP

7nadin3 [17]

Statement 3: tri egc = tri egb
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3 years ago