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Ivenika [448]
4 years ago
6

Exit A five-sided solid has the numbers 1, 2, 3, 4, and 5. What is the probability of rolling two five-sided number solids and g

etting a sum of either a 3 or an 8?
Mathematics
1 answer:
STALIN [3.7K]4 years ago
7 0
To get a sum of 3, you must roll a 1 and a 2. To get a sum of 8, you must roll a 5 and 3, or two 4's.
There are two ways of rolling a 1 and a 2, two ways of rolling a 5 and a 3, and one way to roll two 4's.
There are 5^2=25 possible outcomes when rolling two such solids.

So the probability of getting a sum of 3 or 8 is \dfrac{2+2+1}{25}=\dfrac15
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(-8,3)-(3,-)6 is que 16%. YES
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75.2515 rounded to the tenth place is 80 <br><br> Is this right??
Burka [1]

Answer: False

The tenth place means there will be 1 decimal.

75.2515

because they 5 is left to the 2, and 5 means we round up, the 2 becomes a 3. Therefore, it would equal 75.3

3 0
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Read 2 more answers
OMG can someone please help me solve this problem
klasskru [66]
The correct answer is A.) { - 0.79 ; 1.08 }, because 
<span><span> <span>For </span><span>ax^2 + bx + c = 0</span><span>, the value of </span>x<span> is given by:</span></span>  <span>  ;
a = - 7 ; b = 2 ; c = 6 ; 
b^2 - 4ac = 4 + 168 = 172 ; </span></span>\sqrt{172}≈ 13.11 ;<span><span>
x 1 = ( - 2 + 13.11 ) / (- 14) = 11.11 / ( - 14 ) </span></span>≈<span><span> -0.79 ;
x = 2 = ( - 2 - 13.11 ) / ( - 14 ) = (-15.11) / (- 14  ) </span></span>≈ 1.08 ;<span><span>

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5 0
3 years ago
Last one for this test
Oksanka [162]

Answer:

1) 0

2) 0

3) 0

Explanation:

6 0
3 years ago
Consider independent simple random samples that are taken to test the difference between the means of two populations. The varia
Arturiano [62]

Answer:

d. t distribution with df = 80

Step-by-step explanation:

Assuming this problem:

Consider independent simple random samples that are taken to test the difference between the means of two populations. The variances of the populations are unknown, but are assumed to be equal. The sample sizes of each population are n1 = 37 and n2 = 45. The appropriate distribution to use is the:

a. t distribution with df = 82.

b. t distribution with df = 81.

c. t distribution with df = 41.

d. t distribution with df = 80

Solution to the problem

When we have two independent samples from two normal distributions with equal variances we are assuming that  

\sigma^2_1 =\sigma^2_2 =\sigma^2

And the statistic is given by this formula:

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}

Where t follows a t distribution with n_1+n_2 -2 degrees of freedom and the pooled variance S^2_p is given by this formula:

\S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}

This last one is an unbiased estimator of the common variance \sigma^2

So on this case the degrees of freedom are given by:

df= 37+45-2=80

And the best answer is:

d. t distribution with df = 80

5 0
3 years ago
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