What is the area of 25 pi?

The fourth term of an AP is 13 and the second term is 3 find the common difference<br><br>

kogti [31]

Answer:

5

Step-by-step explanation:

<h3>Given</h3>

a4 = 13
a2 = 3
d = ?

<h3>Solution</h3>

<u>Formula for the terms of AP:</u>

a4 = a + 3d
a2 = a + d

<u>The difference of terms:</u>

a + 3d - (a - d) = 13 - 3
2d = 10
d = 5

3 0

3 years ago

Sasha is making a quilt. Her pattern requires that 1/4 of the pieces of cloth be green. She has 188 pieces of green cloth. If sh

butalik [34]

Answer:

The answer should be 752

Step-by-step explanation:

You turn 1/4 into a decimal and you get 0.25

Divide 188 by .25 and you should get that answer

6 0

3 years ago

How many people were in the sample?<br><br><br><br> 45<br> 90<br> 5<br> 10

Nina [5.8K]

10

The x’s represent just the people, and because the bottom line only represents time it doesn’t affect the amount of people.

4 0

2 years ago

Read 2 more answers

Find the equation of the line containing the points (0,3) and (1,8)

Tema [17]

Answer:

y = 5x + 3

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

To calculate m use the slope formula

m = (y₂ - y₁ ) / (x₂ - x₁ )

with (x₁, y₁ ) = (0, 3) and (x₂, y₂ ) = (1, 8)

m = $\frac{8-3}{1-0}$ = 5

Note the line crosses the y- axis at (0, 3) ⇒ c = 3

y = 5x + 3 ← equation in slope- intercept form

5 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