All categories

3 years ago

Find the perimeter of rectangle PQRS with vertices P(0,0), Q(0,7), R(12,7) & S (12,0)

Mathematics

1 answer:

romanna [79]3 years ago

8 0

P=7+7+12+12P=14+24P=38

See the attachment for diagram

You might be interested in

The temperature dropped 5 °F every hour for 5 hours. What was the total number of degrees the temperature changed in the 5 hours

solong [7]

Answer:

Step-by-step explanation:

you just multiple 5 times 5

4 0

3 years ago

Read 2 more answers

Write an equation (any form) for the quadratic graphed below. Y=

Licemer1 [7]

Answer:

y = -2x^2 - 4x - 1

Step-by-step explanation:

We can see that the graph passes through (-2, -1), (-1, 1) and (0, -1).

Let's solve

ax^2 + bx + c = y

a(-2)^2 + b(-2) + c = -1

4a - 2b + c = -1

a(-1)^2 + b(-1) + c = 1

a - b + c = 1

a0^2 + b0 + c = -1

c = -1

we got c = -1 so we input it into the other 2

4a - 2b - 1 = -1

4a - 2b = 0

2a - b = 0

2a = b

a - b - 1 = 1

a - b = 2

a = b + 2

Let's input b = 2a

a = 2a + 2

-a = 2

a = -2

b = 2a = 2*(-2) = -4

c = -1

y = -2x^2 - 4x - 1

4 0

3 years ago

Solve to find the value of x.

aleksley [76]

135
cross multiply 9*20 then divide by 8

5 0

4 years ago

Read 2 more answers

miss watson runs a distance of 200m in 25 seconds work out the average speed of miss watson in metres per second

Kobotan [32]

Answer:

8 meters per second

Step-by-step explanation:

4 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