Lanzetta Elem. raised 5358....round to 5,000
Lantana Mid. Sch raised 2834....round to 3,000
Atlanta High Sch. raised 4132...round to 4,000

they need 14,000...they have (5,000 + 3,000 + 4,000) = 12,000....so they will need (14,000 - 12,000) = 2,000 more....keep in mind, these are estimates...not exact answers

3 0

3 years ago

Find the measure of angle b. Note that the measure of angle b is acute. Round to the nearest degree.

maks197457 [2]

Answer:

60.75

Step-by-step explanation:

sinA sinB

------- = -----

a b

sin(79)/9 = sinB/8

inverse sin B = 0.87

B = 60.75

6 0

2 years ago

The probability that an American CEO can transact business in a foreign language is .20. Twelve American CEOs are chosen at rand

NNADVOKAT [17]

Answer with Step-by-step explanation:

We are given that

The probability that an American CEO can transact business in foreign language=0.20

The probability than an American CEO can not transact business in foreign language= $1-0.20=0.80$

Total number of American CEOs chosen=12

a. The probability that none can transact business in a foreign language= $12C_0(0.20)^0(0.80)^{12}$

Using binomial theorem $nC_r(1-p)^{n-r}p^r$

The probability that none can transact business in a foreign language= $\frac{12!}{0!(12-0)!}(0.8)^{12}=(0.8)^{12}$

b.The probability that at least two can transact business in a foreign language= $1-P(x=0)-p(x=1)=1-((0.8)^{12}+12C_1(0.8)^{11}(0.2))=1-((0.8)^{12}+12(0.8)^{11}}(0.2))$

c.The probability that all 12 can transact business in a foreign language= $12C_{12}(0.8)^0(0.2)^{12}$

The probability that all 12 can transact business in a foreign language= $\frac{12!}{12!}(0.2)^{12}=(0.2)^{12}$

5 0

2 years ago

Can anyone answer this question ?​

Can anyone answer this question ?