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3 years ago

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For each data set, find the mean, the first and third quartiles, and the interquartiles. Texts per day: 24, 53, 38, 12, 31, 19,

26

1 answer:

kolezko [41]3 years ago

8 0

Mean = 29
Q1 = 19
Q2 = 26
Q3 = 38
IQR = Q3 - Q1 = 38 - 19 = 19

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The expanded form 700,000 + 10,000 + 3,000 + 9 represents the whole number

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Answer:

713,009

Step-by-step explanation:

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3 years ago

Identify the volume and surface area of a sphere in terms of π.

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4 years ago

Starting at 6 a.m. every morning, Matilda receives text messages on her cell phone from her mother, her best friend, and her bro

Dafna1 [17]

Answer:

a) 0.0013 = 0.13% probability that by 7:30 a.m. Mary receives exactly four messages – two of her best friend and two of her mother.

b) $1.6 \times 10^{-9}$ probability that there are no typos in the text messages Matilda receives between 2 p.m. and 5 p.m.

Step-by-step explanation:

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

A) Find the probability that by 7:30 a.m. Mary receives exactly four messages – two of her best friend and two of her mother.

Two from the best friend:

Her best friend sends a message once every 10 minutes.

From 6 to 7:30, there is an hour and a half, that is, 90 minutes, so the mean for her best friend is $\mu = \frac{90}{10} = 9$

Two messages is P(X = 2). So

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 2) = \frac{e^{-9}*9^{2}}{(2)!} = 0.0050$

Two from the mother:

Message every hour = 60 minutes. So $\mu = \frac{90}{60} = 1.5$ . This is P(X = 2).

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 2) = \frac{e^{-1.5}*1.5^{2}}{(2)!} = 0.2510$

Two of her best friend and two of her mother:

Independent events, so the probability of both happening is the multiplication of their separate probabilities.

$p = 0.005*0.251 = 0.0013$

0.0013 = 0.13% probability that by 7:30 a.m. Mary receives exactly four messages – two of her best friend and two of her mother.

B) With a chance of 75% a text message contains a typo independent of the sender. Find the probability that there are no typos in the text messages Matilda receives between 2 p.m. and 5 p.m.

In 3 hours, she is expected to receive:

3*60/10 = 18 messages from her best friend.

3*60/60 = 3 messages from her mother.

3*60/30 = 6 messages from her brother.

In total, 27 messages.

75% probability of a typo, so $\mu = 0.75*27 = 20.25$

This is P(X = 0).

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-20.25}*20.25^{0}}{(0)!} = 1.6 \times 10^{-9}$

$1.6 \times 10^{-9}$ probability that there are no typos in the text messages Matilda receives between 2 p.m. and 5 p.m.

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3 years ago

You lean a ladder 6.7 meters long against a wall. It makes a 63 degree angle with the level ground. How high up is the top of th

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The answer is in the attachment
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4 years ago

Read 2 more answers

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andrezito [222]

<span>4x^3-7x^2-11x+5 </span>÷<span> 4x+5
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12x^2+15x+5
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4x+5
-4x-5
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0</span>

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3 years ago