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3 years ago

The day after Halloween candy was marked down 40% which expression represents the new retail price

Mathematics

1 answer:

nikklg [1K]3 years ago

3 0

Assuming the original price was £1.00
the answer is C) 0.60p

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7 x 28<br> = 7(20 + 8<br> = (7 x 20)+(7<br> = 140 + 56<br> = 196

adell [148]

Answer:B

Step-by-step explanation:

5 0

3 years ago

Please help (its a pic again)

Agata [3.3K]

Answer:

1) v= 280 in^3 sa= 262 in^2

2) v= 80 in^3 sa= 120 in^2

3) v= 36 in^3 sa= 72 in^2

4) v= 240 in^3 sa= 252 in^2

8 0

3 years ago

Circumference of a circle given the radius or diameter? explain why or why not

jek_recluse [69]

Answer: The Radius is the distance from the center outwards. The Diameter goes straight across the circle, through the center. The Circumference is the distance once around the circle.

The formula for calculating the radius is different from calculating the diameter

Find radius

r= diameter/ 2

Radius is half of a diameter

While diameter is

d=2r

Step-by-step explanation:

6 0

3 years ago

What is the area of the object above?

stepan [7]

Answer:

Option C

Step-by-step explanation:

You need to begin by separating the shape into two distinct rectangles. The larger one has dimensions 15 cm * 8cm, while the smaller one has dimensions 7cm * 5cm. Adding these together, you get 15*8+7*5=155 square centimeters, or option C. Hope this helps!

6 0

3 years ago

Read 2 more answers

At a specific point on a highway, vehicles arrive according to a Poisson process. Vehicles are counted in 12 second intervals, a

morpeh [17]

Answer: a) 4.6798, and b) 19.8%.

Step-by-step explanation:

Since we have given that

P(n) = $\dfrac{15}{120}=0.125$

As we know the poisson process, we get that

$P(n)=\dfrac{(\lambda t)^n\times e^{-\lambda t}}{n!}\\\\P(n=0)=0.125=\dfrac{(\lambda \times 14)^0\times e^{-14\lambda}}{0!}\\\\0.125=e^{-14\lambda}\\\\\ln 0.125=-14\lambda\\\\-2.079=-14\lambda\\\\\lambda=\dfrac{2.079}{14}\\\\0.1485=\lambda$

So, for exactly one car would be

P(n=1) is given by

$=\dfrac{(0.1485\times 14)^1\times e^{-0.1485\times 14}}{1!}\\\\=0.2599$

Hence, our required probability is 0.2599.

a. Approximate the number of these intervals in which exactly one car arrives

Number of these intervals in which exactly one car arrives is given by

$0.2599\times 18=4.6798$

We will find the traffic flow q such that

$P(0)=e^{\frac{-qt}{3600}}\\\\0.125=e^{\frac{-18q}{3600}}\\\\0.125=e^{-0.005q}\\\\\ln 0.125=-0.005q\\\\-2.079=-0.005q\\\\q=\dfrac{-2.079}{-0.005}=415.88\ veh/hr$

b. Estimate the percentage of time headways that will be 14 seconds or greater.

so, it becomes,

$P(h\geq 14)=e^{\frac{-qt}{3600}}\\\\P(h\geq 14)=e^{\frac{-415.88\times 14}{3600}}\\\\P(h\geq 14)=0.198\\\\P(h\geq 14)=19.8\%$

Hence, a) 4.6798, and b) 19.8%.

7 0

3 years ago