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Brrunno [24]
2 years ago
5

If MP=6x-5,QR=3x+1, and RN=6, what is QN

Mathematics
1 answer:
stepan [7]2 years ago
7 0

Answer:

12

Step-by-step explanation:

12

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Leno4ka [110]

Answer:The Perimeter of Green Island is 1.6 km.

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3 years ago
Mathematical connection A lighthouse casts a revolving beam of light as far as the pier
Alexandra [31]

The light's coverage area is given mathematically as i.e. Area=\pi d^{2}

Given that a lighthouse casts a revolving beam of light as far as they peir and asked the area is covered.

The revolving beam of light covers the whole space i.e entirely 360°

I.e. it forms a circle-like region

Area of circle = \pi r^{2}{ where r is the radius of circle}

Let's assume the distance traveled by light as "d"

as it forms a circular region, it forms a circle of radius "d"

Mathematical connection i.e. area of the region

⇒area of circular region=\pi d^{2} {where d is the radius of region}

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4 0
1 year ago
P = 2L + 2W<br>How Do I Find W?
tatyana61 [14]
P = 2L + 2W ....subtract 2L from both sides
P - 2L = 2W ...divide both sides by 2
(P - 2L) / 2 = W .....can also be written  like (P/2) - L = W
3 0
2 years ago
Read 2 more answers
The work W required to move a particle from a far distance to within radius r of another particle varies jointly as the product
tangare [24]

Answer:

W = kq1q2 / r

Step-by-step explanation:

W varies jointly as the product of q1 and q2 and inversely as radius r

Product of q1 and q2 = q1q2

W = (k*q1"q2) / r

W = kq1q2 / r

Where,

W = work

q1 = particle 1

q2 = particle 2

r = radius

k = constant of proportionality

The answer is W = kq1q2 / r

5 0
3 years ago
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
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