Simplify the expression: 2(5 + 5q) = What’s the answer

The length of human pregnancies from conception to birth varies according to a distribution that is approximately normal with me

iragen [17]

Answer:

a) 229 and 305 days

b) 229 days or less

c) 305 days or more

Step-by-step explanation:

The Empirical Rule(68-95-99.7 rule) states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 267

Standard deviation = 19

(a) Between what values do the lengths of the middle 95% of all pregnancies fall?_____________and___________days

By the Empirical rule, 95% of all pregnancies fall within 2 standard deviations of the mean.

So

267 - 2*19 = 229 days

to

267 + 2*19 = 305 days

(b) How short are the shortest 2.5% of all pregnancies?______days or less

95% of all pregnancies fall within 2 standard deviations of the mean. The other 5% are more than 2 standard deviations from the mean. Since the distribution is symmetric, 2.5% is more than 2 standard deviations below the mean(shortest 2.5%) and 2.5% is more than 2 standard deviations above the mean(longest 2.5%). So

267 - 2*19 = 229 days

c) How long do the longest 2.5% of pregnancies last?________days or more

Explanation in b)

267 + 2*19 = 305 days

7 0

3 years ago

Patrick paid $4.00 for 5 organic peaches. How much did he pay per peach

leonid [27]

Answer:

80 cents

Step-by-step explanation:

divide the price by 5

4/5=.8

6 0

3 years ago

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Chelsey put for $450 into an account with a simple interest rate of 2.5% when she withdrew the money she had earned a total of $

fgiga [73]

Answer:

5.66 years

Step-by-step explanation:

450 x 1.025^(n) = 450 + 67.50

450 x 1.025^(n) = 517.50

1.025^(n) =(517.50 / 450)

1.025^(n) = 1.15

n = ln(1.15) / ln(1.025)

n = 5.66

6 0

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.

8 0

3 years ago

How to tell if a line is parallel or perpendicular by the equation?

bulgar [2K]

It'll be the opposite reciprocal in front of the X like instead of 3 it will be -1/3 so flip it and change the sign

5 0

3 years ago

Read 2 more answers