To test the claim that the resistance of electric wire can be reduced by more than 0.050 ohm by alloying, thirty two (32) values were obtained for standard wire, and the sample average calculated was equal to 0.136 ohm. In addition, thirty two (32) values were obtained for alloyed wire, and the sample average calculated was equal to 0.083 ohm. The standard deviation resistance for standard wire is known to be 0.004 ohm, whereas the standard deviation resistance for alloyed wire is known to be 0.005. At the level of significance=0.05, does this support the claim?

The z score for this test is z0 =

The p-value for this test is p =

Based on the p-value calculated above, the decision is to fail to reject H0 at a level of significance =0.05 and conclude that the resistance of electric wire can't be reduced by more than 0.05 ohm by alloying. Enter 0 if this statement is FALSE or 1 otherwise:

126.66 m²

step-by-step explanation:

the pertinent area formula is a = πr², where r is the radius.

here, a = π(6.35 m)² = 40.32π, or approximately 126.66 m²

the results from a pre-test for students for the year 2000 and the year 2010 are illustrated in the box plot. what do these results tell us about how students performed on the 29 question pre-test for the two years?

if we compare only the lowest and highest scores between the two years, we might conclude that the students in 2010 did better than the students in 2010. this conclusion seems to follow since the lowest score of 8 in 2010 is greater in value than the lowest score of 6 in 2000. also, the highest score of 28 in 2010 is greater in value than the highest score of 27 in 2000.

but the box portion of the illustration gives us more detailed information. the middle bar in each box shows us that the median score of 20 in 2000 is greater in value than the median score of 17 in 2010. further, we note that the box and whiskers divide the illustration into four pieces. each of these four pieces represents the same portion of students. so, the upper half of the students in 2000 scored in the same score range as the upper one-fourth of the students in 2010, see the illustration at a score of 20.

by considering the upper one-fourth, upper half, and upper three-fourths instead of just the lowest and highest scores, we would conclude that the students as a whole did much better in 2000 than in 2010. we would conclude that as a whole the students in 2010 are less prepared than the students in 2000.

in this section, we discuss box-and-whisker plots and the five key values used in constructing a box-and-whisker plot. the key values are called a five-number summary, which consists of the minimum, first quartile, median, third quartile, and maximum.

step-by-step explanation:

(x) - (y) = (1 + 4i)

1. 180°. the sum of the interior angles of a triangle is 180°, for the equilateral triangle that it has three equals side is 60°+60°+60°=180°.

2. 60°. each interior angle of a equilateral triangle is 60°

3. 120°. the measure of ∠m is 120°, the sum of the angles ∠m and ∠n is equal to 180°. so, in order to find the measure of ∠m, we have ∠m+∠n=180° > ∠m=180°-∠n > ∠m=180°-60°=120°.

4. 360°. the sum of the exterior angles of an equilateral triangle is ∠m+∠r+∠x=120°+120°+120°=360°. the exterior angles of any triangle always add 360°.