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Leviafan [203]
1 year ago
15

A water sprinkler sends water out in a circular pattern. How many feet away from the sprinkler can it spread water if the area f

ormed by the watering pattern is452.16 square​ feet? Use 3.14 for .
Mathematics
2 answers:
Arisa [49]1 year ago
7 0

Answer:

12 feet

Step-by-step explanation:

It asks how many feet away from the sprinkler can water be spread. Therefore, in this scenario, we require the radius.

Area of a circle :

<u><em>Area (circle) = πr²</em></u>

<u><em /></u>

Solving :

⇒ 452.16 = 3.14 × r²

⇒ r² = 452.16 ÷ 3.14

⇒ r = √144

⇒ r = 12 feet

sergey [27]1 year ago
6 0

We need radius

Let that be r

  • πr²=452.16
  • 3.14r²=452.16
  • r²=144
  • r=√144
  • r=12ft
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Answer:

P(5, 0 )

Step-by-step explanation:

Where the line crosses the x- axis the y- coordinate of the point P is zero.

Substitute y = 0 into the equation and solve for x

2x + 3(0) - 10 = 0 , that is

2x - 10 = 0 ( add 10 to both sides )

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3 0
3 years ago
My brother is 4 years younger than my sister and 5 years older than me in 2 years. If I am 15 now how old will my sister be when
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Your sister is 24. Your brother is 20. You are 15. Therefore, your sister will be 48 years old. I’m 87% sure I did this right. Lol, good luck.
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3 years ago
Tim’s savings account pays a simple annual interest rate of 3.5%. Suppose he deposits $5,500 in the account and makes no additio
dimaraw [331]

$6462.5

Simple Interest = (Principal × Rate × Time) / 100

SI = (P×R×T) / 100

P = 5500

R = 3.5

T = 5

[substitute the values into the formula]

SI = (5500 × 3.5 × 5) / 100

SI = 96250 / 100

SI = 962.5

Total = Principal + Simple Interest

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6 0
3 years ago
Suppose a student's score on a standardize test to be a continuous random variable whose distribution follows the Normal curve.
blsea [12.9K]

Answer:

a) Percentage of students scored below 300 is 1.79%.

b) Score puts someone in the 90th percentile is 638.

Step-by-step explanation:

Given : Suppose a student's score on a standardize test to be a continuous random variable whose distribution follows the Normal curve.

(a) If the average test score is 510 with a standard deviation of 100 points.

To find : What percentage of students scored below 300 ?

Solution :

Mean \mu=510,

Standard deviation \sigma=100

Sample mean x=300

Percentage of students scored below 300 is given by,

P(Z\leq \frac{x-\mu}{\sigma})\times 100

=P(Z\leq \frac{300-510}{100})\times 100

=P(Z\leq \frac{-210}{100})\times 100

=P(Z\leq-2.1)\times 100

=0.0179\times 100

=1.79\%

Percentage of students scored below 300 is 1.79%.

(b) What score puts someone in the 90th percentile?

90th percentile is such that,

P(x\leq t)=0.90

Now, P(\frac{x-\mu}{\sigma} < \frac{t-\mu}{\sigma})=0.90

P(Z< \frac{t-\mu}{\sigma})=0.90

\frac{t-\mu}{\sigma}=1.28

\frac{t-510}{100}=1.28

t-510=128

t=128+510

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Score puts someone in the 90th percentile is 638.

5 0
3 years ago
- 5(x - 3) + 4 Completely simplify the expression
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Answer:

Step-by-step explanation:

08– Ao Fatorar a expressão

2+6+9

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2 −9

Leia e responda:

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