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1 year ago

1. Annie and her friends are playing a game called Doubles. In the game, a player

Mathematics

1 answer:

valentina_108 [34]1 year ago

7 0

The sample space is a combination of these two dice as we have shown in the attachment

Die A = {1,2,3,4,5,6}

Die b = {1,2,3,4,5,6}

<h3>What is sample space?</h3>

This is the defined to be the set of all of the possible outcomes that can exist or be possible in a random experiment.

The question has two dice. These dice are made up of numbers from 1 to 6 each.

Die A = {1,2,3,4,5,6}

Die b = {1,2,3,4,5,6}

The sample space would be pairs of two of these two dices. The total number in the sample space would be 36.

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A particular variety of watermelon weighs on average 20.4 pounds with a standard deviation of 1.23 pounds. Consider the sample m

love history [14]

Answer:

a) The expected value of the sample mean weight is 20.4 pounds.

b)The standard deviation of the sample mean weight is 0.123.

c) There is a 14.46% probability the sample mean weight will be less than 20.27.

d) This value is $c = 20.6153$ .

Step-by-step explanation:

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 $\frac{\sigma}{\sqrt{n}}$ .

Problems of normally distributed samples can be 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.

In this problem, we have that:

A particular variety of watermelon weighs on average 20.4 pounds with a standard deviation of 1.23 pounds. This means that $\mu = 20.4, \sigma = 1.23$

Consider the sample mean weight of 100 watermelons of this variety. This means that $n = 100$ .

a. What is the expected value of the sample mean weight? Give an exact answer.

By the Central Limit Theorem, it is the same as the mean of the population. So the expected value of the sample mean weight is 20.4 pounds.

b. What is the standard deviation of the sample mean weight? Give your answer to four decimal places.

By the Central Limit Theorem, that is:

$s = \frac{\sigma}{\sqrt{n}} = \frac{1.23}{\sqrt{100}} = 0.123$

The standard deviation of the sample mean weight is 0.123.

c. What is the approximate probability the sample mean weight will be less than 20.27?

This is the pvalue of Z when $X = 20.27$ .

Since we are working with the sample mean, we use $s$ instead of $\sigma$ in the Z score formula

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

$Z = \frac{20.27 - 20.4}{0.123}$

$Z = -1.06$

$Z = -1.06$ has a pvalue of 0.1446.

This means that there is a 14.46% probability the sample mean weight will be less than 20.27.

d. What is the value c such that the approximate probability the sample mean will be less than c is 0.96?

This is the value of $X = c$ that is in the 96th percentile, that is, it's Z score has a pvalue 0.96.

So we use $Z = 1.75$

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

$1.75 = \frac{c - 20.4}{0.123}$

$c - 20.4 = 0.123*1.75$

$c = 20.6153$

This value is $c = 20.6153$ .

6 0

2 years ago

A farmer owns about 960 acres of land. If his farm is square, what would be the length of each side in miles, to the nearest ten

OLEGan [10]

It is given in the question that

A farmer owns about 960 acres of land. If his farm is square.

First we convert acres to miles square, and the conversion expression is

$1 acre=0.0015625 mi^2$

Therefore,

$960acres=960*0.0015625=1.5 mi^2$

Let the side length of the square be x miles .

And the formula of area of square is

$Area=side^2
\\
1.5=side^2
\\
side= \sqrt{1.5}=1.2 miles$

5 0

3 years ago

Consider rolling two fair dice and observing the number of spots on the resulting upward face of each one. Letting A be the even

Allushta [10]

Answer:

$P(E|A)= \frac{10}{11}$

Step-by-step explanation:

Given

Two rolls of die

$E \to$ one of the outcomes is 6

$A \to$ atleast one is 6

Required

P(E|A)

First, list out the outcome of each

$E = \{(1,6),(2,6),(3,6),(4,6),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5)\}$

$A = \{(1,6),(2,6),(3,6),(4,6),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}$

So:

$P(E|A)= \frac{n(E\ n\ A)}{n(A)}$

Where:

$E\ n\ A = \{(1,6),(2,6),(3,6),(4,6),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5)\}$

$n(E\ n\ A) = 10$

$n(A) = 11$

So:

$P(E|A)= \frac{10}{11}$

7 0

2 years ago

How do I solve this math problem?

nikitadnepr [17]

1,088 if you set it up as a proportion. It would be 17/20 compared to x/1280 multiply 17 by 1280 and then divide by 20.

3 0

3 years ago

Raíz cuadrada de 324

klasskru [66]

Answer:

qie

Step-by-step explanation:

hzhsnshsijxnejeikrnneke

3 0

3 years ago