The national percentage of automobile accident fatalities that are alcohol related is 39%. In a random sample of 96 automobile accident fatalities in the state of Connecticut, 44 were alcohol related. Is this sufficient evidence to say that Connecticut has a higher percentage of alcohol related automobile fatalities?

Step-by-step explanation:

Given that:

There are 96 automobile accident fatalities in Connecticut and 44 were alcohol related.

The null hypothesis and the alternative hypothesis for this study can be computed as:

The sample proportion = 44/96 = 0.45833

Using the Z formula for a single proportion test, we get;

The p-value = 1 -P(Z< 1.37)

The p-value = 1 - 0.9147

The p-value = 0.0853

At the level of significance (∝) 0.05 ;

Since p-value is greater than level of significance (∝);  we accept the null hypothesis and conclude that there is no sufficient evidence to say that higher percentage of alcohol is related to automobile fatalities.

the answer would be 1 1/72

step-by-step explanation:

you can rewrite the equation by separating the parts and then solving individually:

10-9=1

1/8 - 1/9=

find l c d and then combine:

1 1/72