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Ber [7]
3 years ago
15

Police estimate that 25% of drivers drive without their seat belts. If they stop 6 drivers at random, find theprobability that m

ore than 4 are wearing their seat belts.
Mathematics
1 answer:
earnstyle [38]3 years ago
8 0

Answer:

%17.80

Step-by-step explanation:

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What was the average yearly increase at the university of Raleigh for those four years ?
g100num [7]

Answer:

hmm chica, tienes que buscar eso en porque no sé la respuesta a

4 0
2 years ago
1+4=52+5=123+6=218+11=?
Nikolay [14]
1+4=5
2+5=7
3+6=9
8+11=19
hope this helped:)

4 0
3 years ago
Read 2 more answers
What are the measures of < 1 and < 2. Show your work or explain your answer.
VashaNatasha [74]

Answer:

m<1 = 105°

m<2 = 75°

Step-by-step explanation:

Since lines c and d are parallel to each other, therefore:

m<2 = 75° (corresponding angles are congruent)

m<1 + m<2 = 180° (linear angle pair)

Substitute

m<1 + 75° = 180°

Subtract 75 from both sides

m<1 = 180° - 75°

m<1 = 105°

4 0
3 years ago
True or false: 4/8 and 10/16 are equivalent fractions?
lorasvet [3.4K]

Answer:

False

Step-by-step explanation:

Multiply 8 by 2 to create the common denominator (which is 16).

Now multiply 4 by 2 to keep the fraction even.

This gives you the fraction 8/16 which is not equivalent to 10/16.

Another way you could write this inequality:

8/16 < 10/16

4 0
3 years ago
A researcher wants to determine whether the number of minutes adults spend online per day is related to gender. A random sample
suter [353]

Answer:

The expected frequency for the cell E2,2 is = 33

Step-by-step explanation:

The given data is

Gender               | 0-30| 30-60| 60-90| 90+       Totals

Male                    | 23    |    35   |   76    |  46        180

<u> Female               |   31   |    42   |   46    |  16         135    </u>

<u>Totals                    54         77        122    62          315  </u>

<u />

The expected frequency for the cell E2,2 is :

Expected for the (30-60 box for females)   =  Total of (30-6)/ (total )* females

                                                                      = ( 77/315)135= 33

Here p= 77/315 and n= 135

therefor X= pn = 33

χ²=  (33-42)²/33= 2.455  ( for the single value of E2,2=33)

Expected for the (30-60 box for males)   =  Total of (30-6)/ (total )* males

                                                                      = ( 77/315)180= 44

χ²=  (44-42)²/44= 0.09

The critical region is  χ² (0.05) 3 ≥  χ²= 11.34

Let the null and alternate hypothesis be

H0:the number of minutes spent online per day is not related to gender

against

Ha: the number of minutes spent online per day is related to gender

The single value of χ² for E2,2 = 2.45 is less than the critical value of 11.34.  

6 0
2 years ago
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