Solve the equation for all real values of x. sec^2 x + 2 sec x =0

Use technology or a z-score table to answer the question.

AlexFokin [52]

Answer:

D. 100

Step-by-step explanation:

1. Find the z-score: 165-176/11.4 = -1.2

2. Find the percentage of the z-score. Here it's 16.85%

3. Because we are searching for the amount of members that weigh MORE than 165 pounds and not LESS, we need to find the greater percentage by subtracting 16.85% from 100% which is 83.15%. This is the percentage of people that weigh more than 165 pounds.

4. Multiply the percentage by the number of members: 83.15%*120 = 99.78

5. Round it to a whole number because someone can't be 0.78 of a person. You get 100 members.

Also, I took the test :)

7 0

3 years ago

Given the function f(x) = 2(3)x, Section A is from x = 0 to x = 1 and Section B is from x = 2 to x = 3.

jeyben [28]

f(0) = 2(3)(0) = 0 ⇒ (0, 0)

f(1) = 2(3)(1) = 6 ⇒ (1, 6)

rate of change from x = 0 to x = 1

$\frac{y2 - y1}{x2 - x1}$ = $\frac{6 - 0}{1 - 0}$ = $\frac{6}{1}$ = 6

******************************************

f(2) = 2(3)(2) = 12 ⇒ (2, 12)

f(3) = 2(3)(3) = 18 ⇒ (3, 18)

rate of change from x = 2 to x = 3

$\frac{y2 - y1}{x2 - x1}$ = $\frac{18 - 12}{3 - 2}$ = $\frac{6}{1}$ = 6

*****************************************

Since it is the same line, they will have the same slope (aka rate of change).

5 0

4 years ago

On average, Tanner has noticed that 22 trains pass by his house daily (24 hours) on the nearby train tracks. What is the probabi

blondinia [14]

Answer:

0.11069

Step-by-step explanation:

We will assume that the trains pass by his house following a uniform distribution with values between 0 and 24. The probability of a train passing on a 9-hour time period is 9/24 = 3/8 = 0.375. Lets call Y the amount of trains passing by his house during that 9-hour period. Y follows a Binomail distribution with parameters 22 and 0.375.

P(Y ≤ 5) = P(Y = 0) + P(Y=1) + P(Y=2) + P(Y=3) + P(Y=4) + P(Y=5) =

${22 \choose 0} * 0.375^0*(1-0.375)^{22} + {22 \choose 1}*0.375^1*(1-0.375)^{21} +\\\\{22 \choose 2} * 0.375^2*(1-0.375)^{20} + {22 \choose 3}*0.375^3*(1-0.375)^{19} + \\{22 \choose 4} * 0.375^4*(1-0.375)^{18} + {22 \choose 5}*0.375^5*(1-0.375)^{17} = 0.11069$

I hope that works for you!

4 0

3 years ago

Ms kaimal prints out address labels on 3 sheets of paper each sheet is entirely covered in labels each label takes up 1/12 of a

RoseWind [281]

Answer:

Labels did she print is 36.

Step-by-step explanation:

Given : Ms kaimal prints out address labels on 3 sheets of paper each sheet is entirely covered in labels each label takes up $\frac{1}{12}$ of a sheet.