Which of the following are remote interior angles of <6? Check all that apply.

The sum of the angle measures of any triangle is 180 degrees. suppose that one angle in a triangle has a degree measure of −3x5

salantis [7]

If we let m₁, m₂, and m₃ be the measures of the angles of the triangle, the equation that would relate them to each other is,

m₁ + m₂ + m₃ = 180

Given the measures of the first two angles, the measure of the third angle is calculated through the equation,

m₃ = 180 - (m₁ + m₂)

Substituting the known expressions,

m₃ = 180 - (-3x⁵ + 2x²)

Simplifying,

 m₃ = 180 + 3x⁵ - 2x²

7 0

3 years ago

For which inequality is {x | x R, x > 6} the solution set?

CaHeK987 [17]

Answer:

C

Step-by-step explanation:

B: 2X + 12 > 3

2x > -9

x> -9/2

C: 2X - 5 > 7

2x > 12

x > 6

D: 2x - 6 > 24

2x > 30

x > 15

Hope this helps ^-^

6 0

3 years ago

A baseball is thrown into the air from the top of a 224-foot tall building. The baseball's approximate height over time can be r

lianna [129]

H(t) = 0 for t = 7 and for t = -2.

It is reasonable for the time to be positive, 7 seconds.

(The offered answers all appear to be negative, so none are reasonable.)

8 0

3 years ago

Read 2 more answers

Can someone explain how to do this step by step? and give the Formula

scoundrel [369]

It is 0.7p= p-250. 0.7p represents the tv with a 30% discount and p-250 represents the tv with a $250 discount.

3 0

3 years ago

A pen company averages 1.2 defective pens per carton produced (200 pens). The number of defects per carton is Poisson distribute

nlexa [21]

Answer:

a. P(x = 0 | λ = 1.2) = 0.301

b. P(x ≥ 8 | λ = 1.2) = 0.000

c. P(x > 5 | λ = 1.2) = 0.002

Step-by-step explanation:

If the number of defects per carton is Poisson distributed, with parameter 1.2 pens/carton, we can model the probability of k defects as:

$P(k)=\frac{\lambda^{k}e^{-\lambda}}{k!}= \frac{1.2^{k}\cdot e^{-1.2}}{k!}$

a. What is the probability of selecting a carton and finding no defective pens?

This happens for k=0, so the probability is:

$P(0)=\frac{1.2^{0}\cdot e^{-1.2}}{0!}=e^{-1.2}=0.301$

b. What is the probability of finding eight or more defective pens in a carton?

This can be calculated as one minus the probablity of having 7 or less defective pens.

$P(k\geq8)=1-P(k$

$P(0)=1.2^{0} \cdot e^{-1.2}/0!=1*0.3012/1=0.301\\\\P(1)=1.2^{1} \cdot e^{-1.2}/1!=1*0.3012/1=0.361\\\\P(2)=1.2^{2} \cdot e^{-1.2}/2!=1*0.3012/2=0.217\\\\P(3)=1.2^{3} \cdot e^{-1.2}/3!=2*0.3012/6=0.087\\\\P(4)=1.2^{4} \cdot e^{-1.2}/4!=2*0.3012/24=0.026\\\\P(5)=1.2^{5} \cdot e^{-1.2}/5!=2*0.3012/120=0.006\\\\P(6)=1.2^{6} \cdot e^{-1.2}/6!=3*0.3012/720=0.001\\\\P(7)=1.2^{7} \cdot e^{-1.2}/7!=4*0.3012/5040=0\\\\$

$P(k$

c. Suppose a purchaser of these pens will quit buying from the company if a carton contains more than five defective pens. What is the probability that a carton contains more than five defective pens?

We can calculate this as we did the previous question, but for k=5.

$P(k>5)=1-P(k\leq5)=1-\sum_{k=0}^5P(k)\\\\P(k>5)=1-(0.301+0.361+0.217+0.087+0.026+0.006)\\\\P(k>5)=1-0.998=0.002$

5 0

3 years ago