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docker41 [41]
3 years ago
5

I tried to figure the problem out and I didn't know how to do it, can you please help me?

Mathematics
1 answer:
horrorfan [7]3 years ago
6 0

Answer:

i think b would be the best choice.

Step-by-step explanation:

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Find the GCF
Kaylis [27]
Greatest Common Factor. let's go over that. GCF of 6 and 12 would be 6 because 6 goes into 6, 1 times and 6 goes into 12, 2 times. . and it is the largest. so first you see if the lower number can go directly into the higher number. So 28a^2 b^2 c^2, is made of 14abc× 2abc. so you can say 14abc goes into 28a2b2c2, 2abc times. therefore 14abc is the largest factor.
7 0
3 years ago
Engineers must consider the breadths of male heads when designing helmets. The company researchers have determined that the popu
tia_tia [17]

Answer:

The minimum head breadth that will fit the clientele is 4.4 inches.

The maximum head breadth that will fit the clientele is 7.8 inches.

Step-by-step explanation:

Let <em>X</em> = head breadths of men that is considered for the helmets.

The random variable <em>X</em> is normally distributed with mean, <em>μ</em> = 6.1 and standard deviation, <em>σ</em> = 1.

To compute the probability of a normal distribution we first need to convert the raw scores to <em>z</em>-scores using the formula:

z=\frac{x-\mu}{\sigma}

It is provided that the helmets will be designed to fit all men except those with head breadths that are in the smallest 4.3% or largest 4.3%.

Compute the minimum head breadth that will fit the clientele as follows:

P (X < x) = 0.043

⇒ P (Z < z) = 0.043

The value of <em>z</em> for this probability is:

<em>z</em> = -1.717

*Use a <em>z</em>-table.

Compute the value of <em>x</em> as follows:

z=\frac{x-\mu}{\sigma}\\-1.717=\frac{x-6.1}{1}\\x=6.1-(1.717\times 1)\\x=4.383\\x\approx4.4

Thus, the minimum head breadth that will fit the clientele is 4.4 inches.

Compute the maximum head breadth that will fit the clientele as follows:

P (X > x) = 0.043

⇒ P (Z > z) = 0.043

⇒ P (Z < z) = 1 - 0.043

⇒ P (Z < z) = 0.957

The value of <em>z</em> for this probability is:

<em>z</em> = 1.717

*Use a <em>z</em>-table.

Compute the value of <em>x</em> as follows:

z=\frac{x-\mu}{\sigma}\\1.717=\frac{x-6.1}{1}\\x=6.1+(1.717\times 1)\\x=7.817\\x\approx7.8

Thus, the maximum head breadth that will fit the clientele is 7.8 inches.

5 0
3 years ago
Nine minus 3 and 3/8
andriy [413]

Answer:6

Step-by-step explanation:

3 0
3 years ago
What is the mean, median, mode, and range from this data?
34kurt

Answer:

mean = 4

medin = violet

mode = light blue

range = 4

Step-by-step explanation:

mean = sum/no

mode = largest number

median = (n+1)/2th item

rnge = largest - smllest

6 0
3 years ago
Dos cuadrados de lado
Kaylis [27]

La franja amarilla del rectángulo tiene un área de 30 centímetros cuadrados.

<h3>¿Cuál es el área de la franja amarilla del rectángulo?</h3>

En este problema tenemos un rectángulo formado por dos cuadrados que se traslapan uno al otro. La franja amarilla es el área en la que los cuadrados se traslapan. La anchura del rectángulo es descrita por la siguiente ecuación:

(10 - x) + 2 · x = 17

Donde x se mide en centímetros.

A continuación, despejamos x en la ecuación descrita:

10 + x = 17

x = 7

Ahora, el área de la franja amarilla se determina mediante la fórmula de area de un rectángulo:

A = b · h

Donde:

  • b - Base del rectángulo, en centímetros.
  • h - Altura del rectángulo, en centímetros.
  • A - Área del rectángulo, en centímetros cuadrados.

A = (10 - 7) · 10

A = 3 · 10

A = 30

El área de la franja amarilla del rectángulo es igual a 30 centímetros cuadrados.

Para aprender más sobre áreas de rectángulos: brainly.com/question/23058403

#SPJ1

8 0
1 year ago
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