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3 years ago

1, 1, 4, 8, 9 , ?, 16, 64

Mathematics

1 answer:

Elden [556K]3 years ago

8 0

Answer:

Missing number ⇒ 27

Step-by-step explanation:

Given sequence is:

1, 1, 4, 8, 9 , ?, 16, 64

We need to find the missing number of the sequence.

Solution:

We can see the series consist numbers 1, 2, 3, 4, 5.. of power 2 and 3 each respectively. So, the series is $1^{2},1^{3},2^{2},2^{3},3^{2},3^{3}.....$ such that

$1 = 1^{2}\\1=1^{3}\\ 4=2^{2}\\\ 8=2^{3}\\\ 9=3^{2}\\\ 27=3^{3}------missing \ number\\\ 16=4^{2}\\\ 64=4^{3}$

Therefore, missing number is $3^{3}= 27$

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3 years ago

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

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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