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Crazy boy [7]
2 years ago
12

What is the probability of rolling a number less than

Mathematics
1 answer:
sergeinik [125]2 years ago
4 0

Answer:

The correct answer is D. 2/11.

Step-by-step explanation:

To determine what is the probability of rolling a number less than or equal to 8 with the sum of two dice, given that at least one of the dice must show a 6, the following calculation should be performed:

6 + 1 = 7

6 + 2 = 8

2/11 = X

Therefore, the probability of rolling a number less than

or equal to 8 with the sum of two dice, given that at least one of the dice must show a 6, is 2/11.

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Please show work 7r + 21 = 49r
gulaghasi [49]

Answer:

1/2 = r

Step-by-step explanation:

7r + 21 = 49r

Subtract 7r from each side

7r-7r + 21 = 49r-7r

21 = 42r

Divide each side by 2

21/42 = 42r/42

1/2 = r

5 0
2 years ago
Read 2 more answers
An Epson inkjet printer ad advertises that the black ink cartridge will provide enough ink for an average of 245 pages. Assume t
Neko [114]

Answer:

35.2% probability that the sample mean will be 246 pages or more

Step-by-step explanation:

To solve this question, we need to understand 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 normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

\mu = 245 \sigma = 15, n = 33, s = \frac{15}{\sqrt{33}} = 2.61

What the probability that the sample mean will be 246 pages or more?

This is 1 subtracted by the pvalue of Z when X = 246. So

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

By the Central Limit Theorem

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

Z = \frac{246 - 245}{2.61}

Z = 0.38

Z = 0.38 has a pvalue of 0.6480.

1 - 0.6480 = 0.3520

35.2% probability that the sample mean will be 246 pages or more

4 0
3 years ago
Factor the expression below:<br><br> 17x –34
abruzzese [7]

Answer:

Step-by-step explanation:

17(x-2)

8 0
2 years ago
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Please help! Will give brainliest answer
sergiy2304 [10]
You answer is B because supplementary is 180°
3 0
3 years ago
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Solve the equation f(x)=2x+7 ​
bija089 [108]

Answer:

2

Step-by-step explanation:

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3 years ago
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