Na gallop poll of 1012 randomly selected adults, 9% said that cloning of humans should be allowed. we are going to use a .05 significance level to test the claim that less than 10% of all adults say that cloning of humans should be allowed. the test statistic for this test is -1.06. what is the p-value for this test statistic?

0.1446

Step-by-step explanation:

Let p be the population proportion of adults say that cloning of humans should be allowed.

As per given , the appropriate set of hypothesis would be :-

Since is left-tailed and sample size is large(n=1012) , so we perform left-tailed z-test.

Given : The test statistic for this test is -1.06.

P-value for left tailed test,

P(z<-1.06)=1-P(z<1.06)        [∵P(Z<-z)=1-P(Z<z]

=          (Using z-value table.)

Hence, the p-value for this test statistic= 0.1446

7! = 7×6×5×4×3×2

3! = 3× 2

as 7! is multiple of 3! because 7! contains 3! that is 3× 2 as factor

so 7! /(3! ) = 7× 6× 5× 4

= 42 × 20

= 840

its b) 840✌✌✌

dhruv⭐

m∠a is 90°

m∠b is 40°