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lorasvet [3.4K]
3 years ago
15

PLEASE MATH HELP 20 points PLEASE!!!!

Mathematics
1 answer:
blsea [12.9K]3 years ago
7 0
Y = 18( 0.58) ^t

It would be written as :
Y = 18(0.11) ^ (t/4)
Y = 18 (0.11 ^ (1/4)) ^ t
But
0.11 ^ (1/4) = 0.5759

Plug the value

Y = 18 (0.5759) ^ t

Two decimal point :

Y = 18 (0.58) ^ t
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Undefined term in geometry
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Time spent using​ e-mail per session is normally​ distributed, with mu equals 11 minutes and sigma equals 3 minutes. Assume that
liq [111]

Answer:

a) 0.259

b) 0.297

c) 0.497

Step-by-step explanation:

To solve this problem, it is important to know the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 11, \sigma = 3

a. If you select a random sample of 25 ​sessions, what is the probability that the sample mean is between 10.8 and 11.2 ​minutes?

Here we have that n = 25, s = \frac{3}{\sqrt{25}} = 0.6

This probability is the pvalue of Z when X = 11.2 subtracted by the pvalue of Z when X = 10.8.

X = 11.2

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{11.2 - 11}{0.6}

Z = 0.33

Z = 0.33 has a pvalue of 0.6293.

X = 10.8

Z = \frac{X - \mu}{s}

Z = \frac{10.8 - 11}{0.6}

Z = -0.33

Z = -0.33 has a pvalue of 0.3707.

0.6293 - 0.3707 = 0.2586

0.259 probability, rounded to three decimal places.

b. If you select a random sample of 25 ​sessions, what is the probability that the sample mean is between 10.5 and 11 ​minutes?

Subtraction of the pvalue of Z when X = 11 subtracted by the pvalue of Z when X = 10.5. So

X = 11

Z = \frac{X - \mu}{s}

Z = \frac{11 - 11}{0.6}

Z = 0

Z = 0 has a pvalue of 0.5.

X = 10.5

Z = \frac{X - \mu}{s}

Z = \frac{10.5 - 11}{0.6}

Z = -0.83

Z = -0.83 has a pvalue of 0.2033.

0.5 - 0.2033 = 0.2967

0.297, rounded to three decimal places.

c. If you select a random sample of 100 ​sessions, what is the probability that the sample mean is between 10.8 and 11.2 ​minutes?

Here we have that n = 100, s = \frac{3}{\sqrt{100}} = 0.3

This probability is the pvalue of Z when X = 11.2 subtracted by the pvalue of Z when X = 10.8.

X = 11.2

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{11.2 - 11}{0.3}

Z = 0.67

Z = 0.67 has a pvalue of 0.7486.

X = 10.8

Z = \frac{X - \mu}{s}

Z = \frac{10.8 - 11}{0.3}

Z = -0.67

Z = -0.67 has a pvalue of 0.2514.

0.7486 - 0.2514 = 0.4972

0.497, rounded to three decimal places.

5 0
3 years ago
which expression is equivalent to (^3sqrtm)^7 in exponential form? a) 7 m/3 b) 3 m/7 c) m 3/7 d) m 7/3
Komok [63]
(\sqrt[3]{m})^{7} = m^{ \frac{7}{3}}
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The area of a triangle is related to the base and height of the triangle by the formula a = bh/2. Rewrite this equation with b s
SCORPION-xisa [38]

Answer: b = 2a/h

Step-by-step explanation:

a = bh/2

multiply each side by 2

2a = bh

divide each side by h

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PLEASE HELP WITH THE PROBLEM ON THE ATTACHED FILE
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Step-by-step explanation:

First u have to do the work of bracket and the solve

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2 years ago
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