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larisa86 [58]
2 years ago
9

State 5.126 to two decimal places​

Mathematics
1 answer:
Kryger [21]2 years ago
7 0

Answer:

5.13

Step-by-step explanation:

Look at the number at the end, and if it's 5 or more, round up. the 6 is greater than or equal to 5, so you round up, dropping the six and making the 2 into a 3, giving you 5.13.

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Find the area of the circle and the sector.
Georgia [21]

Area of a circle is PI x r^2:

Area of circle = 3.14 x 13^2 = 530.66 square meters.

To find the area of the sector multiply the area of the circle by the fraction of the sector

Area of sector = 530.66 x 150/360 = 221.108 square meters ( Round the answer as needed).

6 0
3 years ago
The picture shows a feeding trough that is shaped like a right prism.
ollegr [7]
A)We have been given →

Hieght of trough = 5m

length of trough = 6m

Breadth of trough = 4m

so , As the top is opened for feeding animals.

Area of base = length× Breadth

= (6×4) m²
= 24 m²

Area of Front and back face = Hieght × length

= (5×6) m²
= 30m²

Area of side faces = hieght × Breadth

= (5×4) m²
= 20 m²
____________________________________

B)Total Surface area of trough will be given by formula - 2(lb+bh+hl)

= 2{(6×4)+(4×5)+(5×6)}

= 2{24+20+30}

= 2×74

= 148 m²

As we have been told top is open , so we will deduct top from Total surface Area to obtain Surface area →

(148-24) m²

(Area of top will be same as base area , that is 24m² as we have solved in A)

= 124m²

Answer→ Surface area of trough is 124m²
___________________________________
5 0
3 years ago
A Geiger counter counts the number of alpha particles from radioactive material. Over a long period of time, an average of 14 pa
UkoKoshka [18]

Answer:

0.2081 = 20.81% probability that at least one particle arrives in a particular one second period.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

x is the number of sucesses

e = 2.71828 is the Euler number

\mu is the mean in the given interval.

Over a long period of time, an average of 14 particles per minute occurs. Assume the arrival of particles at the counter follows a Poisson distribution. Find the probability that at least one particle arrives in a particular one second period.

Each minute has 60 seconds, so \mu = \frac{14}{60} = 0.2333

Either no particle arrives, or at least one does. The sum of the probabilities of these events is decimal 1. So

P(X = 0) + P(X \geq 1) = 1

We want P(X \geq 1). So

P(X \geq 1) = 1 - P(X = 0)

In which

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

P(X = 0) = \frac{e^{-0.2333}*(0.2333)^{0}}{(0)!} = 0.7919

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.7919 = 0.2081

0.2081 = 20.81% probability that at least one particle arrives in a particular one second period.

8 0
3 years ago
Best buy sells iPhones amd Samsung. Thy have 81 phones in the store. If the number of iphones is twice the number of Samsung's h
seropon [69]

Answer:

54  iphones ,27 samsungs

81/3=27

27x2=54

81-54=27

5 0
2 years ago
12a^3b^2 +18a²b^2 – 12ab^2<br> Factor completely
AleksAgata [21]

The factorization of 12a^3b^2 +18a²b^2 – 12ab^2 is 6 a b^{2}(a+2)(2 a-1)

<u>Solution:</u>

Given, expression is 12 a^{3} b^{2}+18 a^{2} b^{2}-12 a b^{2}

We have to factorize the given expression completely.

Now, take the expression

12 a^{3} b^{2}+18 a^{2} b^{2}-12 a b^{2}

Taking b^2 as common term,

b^{2}\left(12 a^{3}+18 a^{2}-12 a\right)

Taking "a" as common term,

b^{2}\left(a\left(12 a^{2}+18 a-12\right)\right)

Taking "6" as common term,

b^{2}\left(a\left(6\left(2 a^{2}+3 a-2\right)\right)\right)

Splitting "3a" as "4a - a" we get,

b^{2}\left(a\left(6\left(2 a^{2}+4 a-a-2\right)\right)\right)

\begin{array}{l}{b^{2}(a(6(2 a(a+2)-1(a+2))))} \\\\ {b^{2}(a(6((a+2) \times(2 a-1))))} \\\\ {6 a b^{2}(a+2)(2 a-1)}\end{array}

Hence, the factored form of given expression is 6 a b^{2}(a+2)(2 a-1)

4 0
2 years ago
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