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3 years ago

What's an area model

1 answer:

4 0

Area model is where the length and width are put together using either multiplication, percentage or fractions to find the size of an area.

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Please help!!! I'll mark you as BRAINLIEST and give you 10 points!!!!!

zheka24 [161]

Answer:

Step-by-step explanation:

Okay so the shape has an area of 60 square inches. The formula for the area of a triangle is A = b*h where A = area, b = base, and h = height

So we are given the value for A (60)

We are given the value for height (10)

So we just need to find the value of the base

So let's set this up as an equation. Input the given values into the area for triangle formula

A = B*h

60 = B * 10

So we need to solve for B. To do that we need to get B alone on one side of the = sign. B is being multiplied by 10 so do the opposite and divide both sides by 10

60 divided by 10 = 6

10 divided by 10 = 1 (don't actually put a 1 down, just drop it)

So we are left with

6 = B

So the base is 6 square inches. To test, put it in the equation:

A = B*h

60 = 6*10

60 = 60

5 0

3 years ago

The table gives the probabilities that new cars of different models will have brake failure. The car model that is least likely

melamori03 [73]

<span>All the information we have are the probabilities, and what we need is the lowest number: so let's choose the smallest probability among the numbers: 0.0065%, B 0.0037%,C 0.0108%,D 0.0029%, E 0.0145%. The smallest of the numbers is 0.0029% -it starts with two 00s and the number that follows, 2, is smaller than all there others - so the smallest probability is in option D - and the model would be the corresponding model (but we're missing some information here) </span>

3 0

3 years ago

Read 2 more answers

2x + 9x2 + 4x + 4xy – 6x2

Harlamova29_29 [7]

Simplified would be 4xy + 3x^2 + 6x

7 0

3 years ago

Read 2 more answers

Engineers must consider the breadths of male heads when designing helmets. The company researchers have determined that the popu

vitfil [10]

Answer:

a) The minimum head breadth that will fit the clientele = 4.105 inches to 3d.p = 4.1 inches to 1 d.p

b) The maximum head breadth that will fit the clientele = 8.905 inches to 3 d.p = 8.9 inches to 1 d.p

Step-by-step explanation:

This is normal distribution problem.

A normal distribution has all the data points symmetrically distributed around the mean in a bell shape.

For this question, mean = xbar = 6.1 inches

Standard deviation = σ = 1 inch

And we want to find the lowermost 2.3% and uppermost 2.3% of the data distribution.

The minimum head breadth that will fit the clientele has a z-score with probability of 2.3% = 0.023

Let that z-score be z'

That is, P(z ≤ z') = 0.023

Using the table to obtain the value of z'

z' = - 1.995

P(z ≤ - 1.995) = 0.023

But z-score is for any value, x, is that value minus the mean then divided by the standard deviation.

z' = (x - xbar)/σ

- 1.995 = (x - 6.1)/1

x = -1.995 + 6.1 = 4.105 inches

The maximum head breadth that will fit the clientele has a z-score with probability of 2.3% also = 0.023

Let that z-score be z''

That is, P(z ≥ z'') = 0.023

Using the table to obtain the value of z''

P(z ≥ z") = P(z ≤ -z")

- z'' = - 1.995

z" = 1.995

P(z ≥ 1.995) = 0.023

But z-score is for any value, x, is that value minus the mean then divided by the standard deviation.

z'' = (x - xbar)/σ

1.995 = (x - 6.1)/1

x = 1.995 + 6.1 = 8.905 inches

6 0

3 years ago

un solar tiene forma de 3/4 de círculo unido con un triángulo rectángulo , En este solar va a ser utilizado para sembrar clavele

Stolb23 [73]

Answer:

$1092671.191

Step-by-step explanation:

<u>Cálculo de Areas</u>

El área de un círculo de radio r se obtiene con la fórmula:

$A_c=\pi r^2$

El área de un triángulo de base b y altura h, perpendicular a la base es:

$A_t=\frac{bh}{2}$

Si el triángulo es rectángulo, b y h son los catetos

El solar tiene una forma tal como se ilustra en la figura anexa. Para averiguar el costo del abono y del techo, debe calcularse primero el área total a cultivar. Primero se determina el área de 3/4 de círculo de radio 5.9 metros

$A_c=\frac{3}{4}\pi r^2$