A Driver’s Ed program is curious if the time of year has an impact on number of car accidents in the U. S. They assumethat weather may have a significant impact on the ability of drivers to control their vehicles. They take a randomsample of 150 car accidents and record the season each occurred in. They found that 27 occurred in the Spring, 39 inthe Summer, 31 in the Fall, and 53 in the Winter. Required:
Can it be concluded at the 0.05 level of significance that caraccidents are not equally distributed throughout the year?

p-value = 0.0145

Hence, Since p-value ( 0.0145 ) is less than significance level ( 0.05 )

we reject null hypothesis.

Therefore, there is sufficient evidence to conclude that Car accidents are NOT equally distributed throughout the year

Step-by-step explanation:

Given the data in the question;

Hypothesis;

Null hypothesis             : H₀ : Car accidents are equally distributed throughout the year

Alternative hypothesis : Hₐ : Car accidents are NOT equally distributed throughout the year

significance level ∝ = 0.05

x ;

Spring = 27

Summer = 39

Fall = 31

Winter = 53

Test Statistics;

Chi Square = ∑[ (O – E)²/E ]

                         O               E                   (O – E)²/E

Spring               27            37.4                2.94

Summer            39            37.4                0.06

Fall                    31             37.4                1.1267

Winter               53            37.4                6.4067

Total                 150           150                 10.5334

so; z = ∑[ (O – E)²/E ] = 10.5334

{from table}

p-value = 0.0145

Hence, Since p-value ( 0.0145 ) is less than significance level ( 0.05 )

we reject null hypothesis.

Therefore, there is sufficient evidence to conclude that Car accidents are NOT equally distributed throughout the year

3/4 ÷ 3/4

step-by-step explanation:

basically divide any proper fraction with itself to fit the criteria.

so if we are using 3/4, then:

3/4 ÷ 3/4 = 1

so 1 is the quotient.

and the numerators for both these fractions is 3.   and if we divide the numerators:

3 ÷ 3 = 1

we see that quotient is equal to our original quotient as well.

5=812a+9+3

step 1: simplify both sides of the equation.

5=812a+9+3

5=(812a)+(9+3)(combine like terms)

5=812a+12

5=812a+12

step 2: flip the equation.

812a+12=5

step 3: subtract 12 from both sides.

812a+12−12=5−12

812a=−7

step 4: divide both sides by 812.

812a /812 = -7/812

a= -1/116

so the answer is

a=   -1/116