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nataly862011 [7]
3 years ago
7

Find the probability of rolling a sum of 9 with two dice

Mathematics
1 answer:
DENIUS [597]3 years ago
6 0
It can be either:
1 and 8
2 and 7
3 and 6
4 and 5
You might be interested in
If f(x) = 2x^3– 5x+ 2, then what is the remainder when f(x) is divided by<br> X – 4?
Mashutka [201]

The remainder is 110 when f(x) = 2x^3-5x+ 2 is divided by x-4

Step-by-step explanation:

We need to find the remainder when f(x) = 2x^3-5x+ 2 is divided by x-4

Solving:

The solution is attached in the figure below.

The quotient is: 2x^2+8x+27

The remainder is: 110

Keywords: quotient, division

Learn more about Division and quotient at:

  • brainly.com/question/629998
  • brainly.com/question/13174289
  • brainly.com/question/1648434

#learnwithBrainly

5 0
3 years ago
Can Somebody Please Explain How To Do It? THANKS
sergejj [24]

Hello from MrBillDoesMath!

Answer:

9


Discussion:

Solution 1:

Distance between two points (x1, y1), (x2,y2) is given by the formula

sqrt(  (x1-x2)^2 + (y1-y2)^2).  In our case this becomes

sqrt( (0-0)^2 +  (12 -3)^2 ) =

sqrt( 9^2) =

9.


Solution 2:

Note points D and E both have x = 0 so segment DE is a vertical segment. The length is simply the difference of the y coordinates: 12- 3 = 9 as before.


Thank you,

MrB

6 0
3 years ago
Use the empirical rule to solve the problem. The annual precipitation for one city is normally distributed with a mean of 288 in
nadezda [96]

Answer:

z=-1.99

z=1.99

And if we solve for a we got

a=288 -1.99*3.7=214.4

a=288 +1.99*3.7=295.4

And the limits for this case are: (214.4; 295.4)

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the annual precipitation of a population, and for this case we know the distribution for X is given by:

X \sim N(288,3.7)  

Where \mu=288 and \sigma=3.7

The confidence level is 95.44 and the signficance is 1-0.9544=0.0456 and the value of \alpha/2 =0.0228. And the critical value for this case is z = \pm 1.99

Using this condition we can find the limits

z=-1.99

z=1.99

And if we solve for a we got

a=288 -1.99*3.7=214.4

a=288 +1.99*3.7=295.4

And the limits for this case are: (214.4; 295.4)

8 0
3 years ago
-5⁄2 a + 5 = 25 What Does A =
ruslelena [56]

Answer:

a = -8

Step-by-step explanation:

-5⁄2 a + 5 = 25

Subtract 5 from each side

-5⁄2 a + 5-5 = 25-5

-5/2 a = 20

Multiply each side by -2/5 to isolate a

-2/5 * -5/2 a = -2/5 *20

a = -8

6 0
3 years ago
There are 15 candidates for 4 job positions.
Veronika [31]

Answer:

32760 ways

Step-by-step explanation:

Given

Number of Candidates = 15

Job Positions = 4

Required:

Number of outcomes

This question represent selection; i.e. selecting candidates for job positions;

This question can be solved in any of the following two ways

Method 1.

The first candidate can be chosen from any of the 15 candidates

The second candidate can be chosen from any of the remaining 14 candidates

The third candidate can be chosen from any of the remaining 13 candidates

The fourth candidate can be chosen from any of the remaining 12 candidates

Total Possible Selection = 15 * 14 * 13 * 12

<em>Total Possible Selection = 32760 ways</em>

<em></em>

Method 2:

This can be solved using permutation method which goes thus;

nPr = \frac{n!}{(n-r)!}

Where n = 15 and r = 4

So;

nPr = \frac{n!}{(n-r)!} becomes

15P4 = \frac{15!}{(15-4)!}

15P4 = \frac{15!}{11!}

15P4 = \frac{15*14*13*12*11!}{11!}

15P4 = 15*14*13*12

15C4 = 32760

<em>Hence, there are 32760 ways</em>

8 0
3 years ago
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