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gayaneshka [121]
3 years ago
7

Find the experimental probability Roll dice 1,3,3,4,4 P(1)= I’ll give brainleist

Mathematics
2 answers:
MaRussiya [10]3 years ago
8 0

Answer:

1/5

Step-by-step explanation:

-Dominant- [34]3 years ago
5 0

Answer:

\mathrm{P(1)\:}=\frac{1}{5}

Step-by-step explanation:

There are simply 5 possible values in the given set. Out of these, only one of these is the number 1. Therefore, the probability a 1 is drawn (P(1)) is \fbox{$\frac{1}{5}$}.

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Which inequality is a true statement?
lubasha [3.4K]

9514 1404 393

Answer:

  • -3 ≤ -1
  • -3 < -1

Step-by-step explanation:

Negative three is less than (<) negative one, so any statement saying otherwise is incorrect. Correct choices are ...

  -3 ≤ -1

  -3 < -1

7 0
3 years ago
If the m ZABC = 60°, find mZABD.
Vinil7 [7]

Answer:

39.5

Step-by-step explanation:

angle ABC=60=(5x+2)+(3x-2)

60=8x

x=7.5

Angle ABD=(5x+2),(sub x=7.5)

                  =(5*7.5+2)

                  =39.5

7 0
4 years ago
Nurse Jackson examines immunization records for this past winter at the school where she works. Got flu Didn't get flu Total Vac
Harman [31]

Answer:

140

Step-by-step explanation:

ap3x

6 0
3 years ago
Engineers measure angles in gradients, which are smaller than degrees. The table shows the conversion of some angle measures in
Schach [20]

Answer:

1.11

Step-by-step explanation:

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5 0
4 years ago
The proportion of households in a region that do some or all of their banking on the Internet is 0.31. In a random sample of 100
Alenkasestr [34]

Answer:

Approximate probability that the number of households that use the Internet for banking in a sample of 1000 is less than or equal to 130 is less than 0.0005% .

Step-by-step explanation:

We are given that let X be the number that do some or all of their banking on the Internet.

Also; Mean, \mu = 310/1000 or 0.31   and  Standard deviation, \sigma = 14.63/1000 = 0.01463 .

We know that Z = \frac{X-\mu}{\sigma} ~ N(0,1)

Probability that the number of households that use the Internet for banking in a sample of 1000 is less than or equal to 130 is given by P(X <= 130/1000);

 P(X <=0.13) = P( \frac{X-\mu}{\sigma}  <= \frac{0.13-0.31}{0.01463} ) = P(Z <= -12.303) = P(Z > 12.303)

Since this value is not represented in the z table as the value is very high and z table is limited to x = 4.4172.

So, after seeing the table we can say that this probability is approximately less than 0.0005% .

4 0
4 years ago
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