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ella [17]
3 years ago
13

Morgan is playing a board game that requires three standard dice to be thrown at one time. Each die has six sides, with one of t

he numbers 1 through 6 on each side. She has one throw of the dice left, and she needs a 17 to win the game. What is the probability that Morgan wins the game (order matters)?
Mathematics
1 answer:
PIT_PIT [208]3 years ago
8 0
The probability of having a 17 is the number of ways of having 17 divided by the number of different possible outcomes.

Ways of having 17:

Die 1     Die 2     Die 3    sum
6            6            5          6+6+5 = 17
6            5            6          6+5+6 = 17
5            6            6          5+6+6 = 17

Those are the only 3 possibilities of having 17 (you can only loose 1 point).

The total number of outcomes are 6*6*6 = 216

So the chances (probability) to win is 3 / 216 = 1/72 = 0.0139 
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TV has a listed price of 623.98 if the sales tax rate is 8.5 find the total cost of the TV with the sales tax included​
Anastasy [175]

Answer:

$677.02

Step-by-step explanation:

623.98 increase 8.5% =

623.98 × (1 + 8.5%) = 623.98 × (1 + 0.085) = 677.0183

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Awnser in one/two step equations
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-6=-7+x

Make the variable to the left hand side and change it’s sign.

-6-x=-7

Move the constant to the right hand side and change its sign.

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2 years ago
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1. A food safety guideline is that the mercury in fish should be below 1 part per million​ (ppm). Listed below are the amounts o
Sauron [17]

Answer:

The confidence interval estimate of the population mean is :

(0.61 ppm, 0.90 ppm)

The correct option is (A).

Step-by-step explanation:

The amounts of mercury​ (ppm) found in tuna sushi sampled at different stores in a major city are:

S = {0.58, 0.82, 0.10, 0.98, 1.27, 0.56, 0.96}

A (100-\alpha )\% confidence interval for the population mean (μ) is an interval estimate of the true value of the mean. This interval has a (100-\alpha )\% probability of consisting the true value of mean.

⇒ Since the population standard deviation is not provided we will use the <em>t</em>-distribution to construct the 99% confidence interval for mean.

⇒ The formula for confidence interval for the population mean is:

                                          \bar x\pm t_{\alpha/2,(n-1)}\times \frac{s}{\sqrt{n}}

Here,

\bar x = sample mean

<em>s </em>= sample standard deviation

<em>n</em> = sample size

t_{\alpha/2,(n-1)} = critical value.

The degrees of freedom for the critical value is, (<em>n</em> - 1) = 7 - 1 = 6.

The significance level is: \alpha =1-Confidence\ level=1-0.99=0.01

The critical value is:

t_{\alpha/2,(n-1)}=t_{0.01/2, 7}=t_{0.05,7}=3.143

**Use the <em>t</em>-table for the critical value.

Compute the sample mean and sample standard deviation as follows:

\bar x=\frac{1}{7}(0.58+ 0.82+ 0.10+ 0.98+ 1.27+ 0.56+ 0.96) =0.753

\int\limits^a_b {x} \, dx s=\sqrt{\frac{\sum (x{i}-\bar x)^{2}}{n-1} } =\sqrt{\frac{1}{6} \times 0.859743} =0.379

The 99% confidence interval for μ is:

x^{2} CI=0.753\pm 3.143\times\frac{0.379}{\sqrt{7}} \\=0.753\pm0.143\\=(0.61, 0.896)\\\approx(0.61, 0.90)

The confidence interval estimate of the population mean is:

(0.61 ppm, 0.90 ppm)

The upper and lower limit of the 99% confidence interval indicates that the true mean value is less than 1 ppm. This implies that there is not too much mercury in tuna sushi

Because it is possible that the mean is not greater than 1 ppm. Also, at least one of the sample values is less than 1 ppm, so at least some of the fish are safe.

Thus, the correct option is (A).

6 0
3 years ago
Peter began his trip at 1 on the number line shown above. He moved 3 positive counts on day one and 8 negative counts on day two
Montano1993 [528]

Answer:

-4

Step-by-step explanation:

1 + 3 = 4

4 - 8 = -4

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2 years ago
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