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2 years ago

A multiple must be greater than or equal to the largest starting number. True or False

Mathematics

1 answer:

podryga [215]2 years ago

8 0

I'll use multiples of 2 and 4 as an example:

Multiples of 2: 2, 4, 6, 8...

Multiples of 4: 4, 8, 12, 16...

The least common multiple in this case is 4. The LCM is always ≥ the largest starting number, which is 4 for this example. Therefore, the statement is true.

<em>Hope this helps! :)</em>

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The yard behind the Cindy's house is rectangular in shape and has a perimeter of 72 feet. If the length l of the yard is 18 feet

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Let the width of the yard be w.
Since the length is 18feet longer, l = w + 18

Perimeter for rectangle = 2(l + w)

2(l + w) = 72

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   2w + 18 = 36

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2 years ago

A random sample of 28 statistics tutorials was selected from the past 5 years and the percentage of students absent from each on

SVEN [57.7K]

Answer:

Step-by-step explanation:

Hello!

X: number of absences per tutorial per student over the past 5 years(percentage)

X≈N(μ;σ²)

You have to construct a 90% to estimate the population mean of the percentage of absences per tutorial of the students over the past 5 years.

The formula for the CI is:

X[bar] ± $Z_{1-\alpha /2}$ * $\frac{S}{\sqrt{n} }$

⇒ The population standard deviation is unknown and since the distribution is approximate, I'll use the estimation of the standard deviation in place of the population parameter.

Number of Absences 13.9 16.4 12.3 13.2 8.4 4.4 10.3 8.8 4.8 10.9 15.9 9.7 4.5 11.5 5.7 10.8 9.7 8.2 10.3 12.2 10.6 16.2 15.2 1.7 11.7 11.9 10.0 12.4

X[bar]= 10.41

S= 3.71

$Z_{1-\alpha /2}= Z_{0.95}= 1.645$

[10.41±1.645* $(\frac{3.71}{\sqrt{28} } )$ ]

[9.26; 11.56]

Using a confidence level of 90% you'd expect that the interval [9.26; 11.56]% contains the value of the population mean of the percentage of absences per tutorial of the students over the past 5 years.

I hope this helps!

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2 years ago