Given that the first spinner has three equal sectors labelled 1, 2 and 3; and the second spinner has equal sectors labelled 3, 4, 5 and 6.

The number of possible outcomes that do not show a 1 on the first spinner is 2 (i.e. the first spinner shows 2 or 3).

The number of possible outcomes that the second spinner show the number 4 is 1 (i.e. the second spinner shows 4)

In probability, the word 'and' goes with multiplication.

Therefore, the number of possible outcomes that do not show a 1 on the first spinner and show the number 4 on the second spinner is given by 2 x 1 = 2 possible outcomes.

i.e. the first spinner shows the number 2 and the second spinner shows the number 4 or the first spinner shows the number 3 and the second spinner shows the number 4.

5 0

3 years ago

Figure ABCD is graphed on a coordinate plane.

Phoenix [80]

1. BC is 1+3=4 (units)

2. AD is 3+5= 8 (units)

3. Draw the altitude CH from point C, as shown in the figure.

4. Then triangle CHD is a right triangle, with hypothenuse CD, and sides HD and CH.

5. $CD^2=HD^2+CH^2$

$CD^2=2^2+4^2=4+16=20$

$CD= \sqrt{20}= \sqrt{4*5}= \sqrt{4} * \sqrt{5}=2 \sqrt{5}= 4.47$

6. AB=CD=4.47 because the trapezoid is isosceles.

7. P= BC+AD+AB+CD=4+8+4.47+4.47=20.9 (units)