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agasfer [191]
3 years ago
12

A reactangular court measures 600 cm wide and 700 cm long find the perimeter?

Mathematics
1 answer:
Tcecarenko [31]3 years ago
4 0

Answer:

2(l+b)

2(600cm+700cm)

2(1300cm)

2600cm

Therefore, the perimeter of the rectangular court is 2600cm.

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kompoz [17]

3x − 2y < 7

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3 years ago
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Faliha runs around a square field of side 70m, while Rameen runs around a rectangular field with length 90m and breadth 60m. Who
inn [45]

\huge\sf\green{Answer :- }

<h2>Given :- </h2><h2 />
  • Side of the square field = 70m

  • Length of the rectangular field = 90m

  • Breadth of the rectangular field = 60m

<h2>To Find :-</h2>

  • Perimeter of the square field

  • Perimeter of the rectangular field

<h2>Solution :-</h2>

Perimeter of the square = 4 × side

By substituting values,

⇒ 4 × 70m

➥ 280m

<h3><u>Hence, </u><u>Faliha</u><u> runs 280m around a Square </u><u>field</u></h3>

Perimeter of the rectangle = 2( length + Breadth)

By substituting values,

⇒ 2( 90m + 60m )

⇒ 2( 150m)

➥ 300m

<h3><u>Hence, Rameen runs 300m around a rectangular field</u></h3>

By comparing perimeters, Rameen runs more than Faliha

Distance runs more by rameen = Distance covered by rameen - distance covered by faliha

By substituting values,

⇒ 300m - 280m

➠ 20m

<h3><u>Hence, </u><u>Rameen</u><u> runs more around a rectangular field by 20</u><u>m</u></h3>

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3 years ago
WILL AWARD BRAINLIEST
Julli [10]
The given is that y varies directly with x:
y = xk
8 = 2k
k = 4

Now we know that the constant is 4. Make a new equation:

y = xk
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4 0
3 years ago
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A​ bank's loan officer rates applicants for credit. The ratings are normally distributed with a mean of 200 and a standard devia
Butoxors [25]

Answer:

Probability that their mean is above 215 is 0.0287.

Step-by-step explanation:

We are given that a bank's loan officer rates applicants for credit. The ratings are normally distributed with a mean of 200 and a standard deviation of 50.

For this, 40 different applicants are randomly​ selected.

<em>Let X = ratings for credit</em>

So, X ~ N(\mu=200,\sigma^{2}=50^{2})

Now, the z score probability distribution for sample mean is given by;

         Z = \frac{X-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \mu = population mean = 200

           \sigma = standard deviation = 50

           \bar X = sample mean

           n = sample of applicants = 40

So, probability that their mean is above 215 is given by = P(\bar X > 215)

    P(\bar X > 215) = P( \frac{X-\mu}{\frac{\sigma}{\sqrt{n} } } > \frac{215-200}{\frac{50}{\sqrt{40} } } ) = P(Z > 1.897) = 1 - P(Z \leq 1.897)

                                                         = 1 - 0.97108 = 0.0287

Therefore, probability that their mean is above 215 is 0.0287.

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3 years ago
Unit 1 Lesson 4 Quiz
Arisa [49]

Answer:

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Step-by-step explanation:

5 0
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