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3 years ago

How to work out the length and the width of the rectangle

Mathematics

1 answer:

wariber [46]3 years ago

3 0

Rectangle:

Define x:

Let the width be x.
width = x
Length = 3x + 4

Find Area:

Area = x(3x + 4)

Square:
Length of square = Length of rectangle = 3x + 4

Area = (3x + 4)²

Area of square is 66cm² more than area of rectangle.
⇒ (3x+ 4)² = x(3x + 4) + 66

Solve x;
(3x+ 4)² = x(3x + 4) + 66

Remove brackets:
9x² + 24x + 16 = 3x² + 4x + 66

Take away 3x² + 4x + 66 from both sides:
6x² + 20x - 50 = 0

Divide by 2:
3x² + 10x - 25 = 0

Factorise:
(3x - 5)(x + 5) = 0

Apply product multiplication property
(3x - 5)= 0 or (x + 5) = 0
x = 5/3 or x = -5 (rejected, since x cannot be negative)

Width x= 5/3 cm
Length = 3x + 4 = 3(5/3) + 4 = 9 cm

Answer: The width is 5/3 cm and length is 9 cm.

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1.) Given : S is the midpoint of RT; PR = PT

garik1379 [7]

Answer: 1.

Given: S is the midpoint of RT

PR=PT

To Prove: PRS=PTS

Step-by-step explanation:

If PR=PT in a triangle PRT, then the triangle is isosceles triangle.

Therefore, ∠PRT=∠PTR (opposite angles of an isosceles triangle are equal)

In ▲PRS and ▲PTS

PR=PT

∠PRT=∠PTR

PS=PS (PS is common)

therefore, ▲PRS≅▲PTS

Therefore, ∠PRS=∠PTS

Hence proved.

Answer: 2.

Given: E is the midpoint of AB and CD.

To Prove: AEC=BED

Step-by-step explanation:

If AB and CD are the diagonals of rectangle, then

diagonals AB=CD, sides AC=BD and AD=BC

In ▲AEC and ▲BED,

∠AEC=∠BED(vertically opposite angles)

AE=BE

AC=BD

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3 years ago

Let z denote a random variable that has a standard normal distribution. Determine each of the probabilities below. (Round all an

Gelneren [198K]

Answer:

(a) P (Z < 2.36) = 0.9909 (b) P (Z > 2.36) = 0.0091

(e) P (-0.77< Z > -0.55) = 0.0705 (f) P (Z > 3) = 0.0014

(g) P (Z > -3.28) = 0.9995 (h) P (Z < 4.98) = 0.9999.

Step-by-step explanation:

Let us consider a random variable, $X \sim N (\mu, \sigma^{2})$ , then $Z=\frac{X-\mu}{\sigma}$ , is a standard normal variate with mean, E (Z) = 0 and Var (Z) = 1. That is, $Z \sim N (0, 1)$ .

In statistics, a standardized score is the number of standard deviations an observation or data point is above the mean. The z-scores are standardized scores.

The distribution of these z-scores is known as the standard normal distribution.

(a)

Compute the value of P (Z < 2.36) as follows:

P (Z < 2.36) = 0.99086

≈ 0.9909

Thus, the value of P (Z < 2.36) is 0.9909.

(b)

Compute the value of P (Z > 2.36) as follows:

P (Z > 2.36) = 1 - P (Z < 2.36)

= 1 - 0.99086

= 0.00914

≈ 0.0091

Thus, the value of P (Z > 2.36) is 0.0091.

(c)

Compute the value of P (Z < -1.22) as follows:

P (Z < -1.22) = 0.11123

≈ 0.1112

Thus, the value of P (Z < -1.22) is 0.1112.

(d)

Compute the value of P (1.13 < Z > 3.35) as follows:

P (1.13 < Z > 3.35) = P (Z < 3.35) - P (Z < 1.13)

= 0.99960 - 0.87076

= 0.12884

≈ 0.1288

Thus, the value of P (1.13 < Z > 3.35) is 0.1288.

(e)

Compute the value of P (-0.77< Z > -0.55) as follows:

P (-0.77< Z > -0.55) = P (Z < -0.55) - P (Z < -0.77)

= 0.29116 - 0.22065

= 0.07051

≈ 0.0705

Thus, the value of P (-0.77< Z > -0.55) is 0.0705.

(f)

Compute the value of P (Z > 3) as follows:

P (Z > 3) = 1 - P (Z < 3)

= 1 - 0.99865

= 0.00135

≈ 0.0014

Thus, the value of P (Z > 3) is 0.0014.

(g)

Compute the value of P (Z > -3.28) as follows:

P (Z > -3.28) = P (Z < 3.28)

= 0.99948

≈ 0.9995

Thus, the value of P (Z > -3.28) is 0.9995.

(h)

Compute the value of P (Z < 4.98) as follows:

P (Z < 4.98) = 0.99999

≈ 0.9999

Thus, the value of P (Z < 4.98) is 0.9999.

**Use the z-table for the probabilities.

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3 years ago