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uranmaximum [27]
3 years ago
14

Find the LCM of 91, 143 and 169​

Mathematics
2 answers:
lidiya [134]3 years ago
8 0

<em>91 = 7×13</em>

<em>143 = 11×13</em>

<em>169 = 13²</em>

<em>LCM = 7×11×13² = 77×169 = 13013</em>

Allushta [10]3 years ago
7 0

Answer:

13013

Step-by-step explanation:

Express each number in terms of the product of prime numbers.

91 = 7 × 13

143 = 11 × 13

169 = 13 × 13

Now consider the maximum number of times each factor occurs for each number.

The factor 7 occurs once

The factor 11 occurs once

The factor 13 is repeated twice ( maximum number of times for 1 number )

Thus LCM = 7 × 11 × 13 × 13 = 13013

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posledela

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Answer:

15.5 inches

Step-by-step explanation:

  • Length of rectangle = 22 inches
  • Perimeter = 75 inches

To calculate : Width of the rectangle.

★ Perimeter of rectangle = 2( l + w )

→ 75 = 2(22 + w)

→ 75 = 44 + 2w

→ 75 – 44 = 2w

→ 31 = 2w

→ 31 ÷ 2 = w

→ <u>1</u><u>5</u><u>.</u><u>5</u><u> </u><u>inches</u><u> </u><u>=</u><u> </u><u>Width</u>

Therefore, width of the rectangle is 15.5 inches.

6 0
3 years ago
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6 0
3 years ago
The heights of a certain type of tree are approximately normally distributed with a mean height p = 5 ft and a standard
arsen [322]

Answer:

A tree with a height of 6.2 ft is 3 standard deviations above the mean

Step-by-step explanation:

⇒ 1^s^t statement: A tree with a height of 5.4 ft is 1 standard deviation below the mean(FALSE)

an X value is found Z standard deviations from the mean mu if:

\frac{X-\mu}{\sigma} = Z

In this case we have:  \mu=5\ ft\sigma=0.4\ ft

We have four different values of X and we must calculate the Z-score for each

For X =5.4\ ft

Z=\frac{X-\mu}{\sigma}\\Z=\frac{5.4-5}{0.4}=1

Therefore, A tree with a height of 5.4 ft is 1 standard deviation above the mean.

⇒2^n^d statement:A tree with a height of 4.6 ft is 1 standard deviation above the mean. (FALSE)

For X =4.6 ft  

Z=\frac{X-\mu}{\sigma}\\Z=\frac{4.6-5}{0.4}=-1

Therefore, a tree with a height of 4.6 ft is 1 standard deviation below the mean .

⇒3^r^d statement:A tree with a height of 5.8 ft is 2.5 standard deviations above the mean (FALSE)

For X =5.8 ft

Z=\frac{X-\mu}{\sigma}\\Z=\frac{5.8-5}{0.4}=2

Therefore, a tree with a height of 5.8 ft is 2 standard deviation above the mean.

⇒4^t^h statement:A tree with a height of 6.2 ft is 3 standard deviations above the mean. (TRUE)

For X =6.2\ ft

Z=\frac{X-\mu}{\sigma}\\Z=\frac{6.2-5}{0.4}=3

Therefore, a tree with a height of 6.2 ft is 3 standard deviations above the mean.

6 0
3 years ago
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