WILL GIVE BRAINLIEST PLEASE HELP

Solve the compound inequality. 3x − 4 > 5 or 1 − 2x ≥ 7 A. x ≤ −3 or x > 3 B. x −3 or x ≤ 3

iris [78.8K]

Answer: B

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

Evaluate the expression. [(3–5)(34)]3

yawa3891 [41]

Answer:

-204

Step-by-step explanation:

(3−5)(34)(3)

=−204

7 0

2 years ago

53% of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the

faltersainse [42]

Answer:

a

$P(X = 5) = 0.2423$

b

$P( X \ge 6 )= 0.4516$

c

$P( X < 4 )= 0.12694$

Step-by-step explanation:

From the question we are told that

The population proportion is p = 0.53

The sample size is n = 10

Generally the distribution of the confidence of US adults in newspapers follows a binomial distribution

i.e

$X \~ \ \ \ B(n , p)$

and the probability distribution function for binomial distribution is

$P(X = x) = ^{n}C_x * p^x * (1- p)^{n-x}$

Here C stands for combination hence we are going to be making use of the combination function in our calculators

Generally the probability that the number of U.S. adults who have very little confidence in newspapers is exactly five is mathematically represented as

$P(X = 5) = ^{10}C_5 * 0.53^5 * (1- 0.53)^{10-5}$

=> $P(X = 5) = 252 * 0.04182 * 0.023$

=> $P(X = 5) = 0.2423$

Generally the probability that the number of U.S. adults who have very little confidence in newspapers is at least six is mathematically represented as

$P( X \ge 6 )= P( X = 6) + P( X = 7) + P( X = 8) + P( X = 9) + P( X = 10)$

=> $P( X \ge 6 )= [^{10}C_6 * [0.53]^6 * (1- 0.53)^{10-6}] + [^{10}C_7 * [0.53]^7 * (1- 0.53)^{10-7}] + [^{10}C_8 * [0.53]^8 * (1- 0.53)^{10-8}] + [^{10}C_9 * [0.53]^9 * (1- 0.53)^{10-9}] + [^{10}C_{10} * [0.53]^{10} * (1- 0.53)^{10-10}]$

=> $P( X \ge 6 )= [0.227] + [0.1464] + [0.0619] + [0.0155] + [0.00082]$

=> $P( X \ge 6 )= 0.4516$

Generally the probability that the number of U.S. adults who have very little confidence in newspapers is less than four is mathematically represented as

$P( X < 4 )= P( X = 3) + P( X = 2) + P( X = 1) + P( X = 0)$

=> $P( X < 4 )= [^{10}C_3 * 0.53^3 * (1- 0.53)^{10-3}] + [^{10}C_2 * 0.53^2 * (1- 0.53)^{10-2}] + [^{10}C_1 * 0.53^1 * (1- 0.53)^{10-1}] + [^{10}C_0 * 0.53^0 * (1- 0.53)^{10-0}]$

=> $P( X < 4 )= 0.09051 + 0.03 + 0.0059 + 0.000526$

=> $P( X < 4 )= 0.12694$

5 0

2 years ago

How many times does X squared minus 4X -12 cross the X axis

kolbaska11 [484]

We can solve this problem using discriminant.

x^2-4x-12's discriminant is

(-4)^2-4*-12=16+48 which is clearly larger than 0

This means that it crosses over the axes 2 times.

In case you don't know what discriminant is, its in equation ax^2+bx+c

the discriminant is b^2-4ac.

If its positive it has 2 crosses with x axis, if negative then 0 crosses, if 0 then 1 cross.

Hope this helped at least a little bit :D

4 0

2 years ago

What is the midpoint of the line segment with endpoints (-5.5, -6.1) and (-0.5,9.1)?

umka21 [38]

(-3, 1.5)....simply add the x coordinates together, divide them by two, and then do the same for the y coordinates.

4 0

3 years ago