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2 years ago

What is the least common multiple of $441$ and $1008$?

2 answers:

6 0

To get the Least Common Multiple (LCM) of 441 and 1008 we need to factor each value first and then we choose all the factors which appear in any column and multiply them:

441: 3 3 7 7
1008: 2 2 2 2 3 3 7
LCM: 2 2 2 2 3 3 7 7
The Least Common Multiple (LCM) is: 2 x 2 x 2 x 2 x 3 x 3 x 7 x 7 = 7056

dolphi86 [110]2 years ago

5 0

Answer:

7056 good luck on this

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Errors in an experimental transmission channel are found when thetransmission is checked by a certifier that detects missing pul

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Answer:

As shown below

Step-by-step explanation:

Given that when X denotes the errors in an experimental transmission channel, when checked by a certifier that detects missing pulses. follows the cumulative density function as given below:

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3 years ago

The amount of money spent on textbooks per year for students is approximately normal.

Ostrovityanka [42]

Answer:a

$336.04 < \mu < 443.96$

The margin of error will increase

The margin of error will decreases

The 99% confidence interval is $0.4107 < p < 0.4293$

Step-by-step explanation:

From the question we are told that

The sample size $n = 19$

The sample mean is $\= x = \$\ 390$

The standard deviation is $\sigma = \$ \ 120$

Given that the confidence level is 95% then the level of significance is mathematically represented as

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha = 0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table

$Z_{\frac{\alpha }{2} } = 1.96$

The margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma}{\sqrt{n} }$

=> $E = 1.96 * \frac{120}{\sqrt{19} }$

=> $E = 53.96$

The 95% confidence interval is

$\= x - E < \mu < \= x + E$

=> $390 - 53.96 < \mu < 390 - 53.96$

=> $336.04 < \mu < 443.96$

When the confidence level increases the $Z_{\frac{\alpha }{2} }$ also increases which increases the margin of error hence the confidence level becomes wider