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3 years ago

Sequence 4,16,36,64,100

2 answers:

7 0

If you're looking for the rule, it's: they're all even squared numbers
i.e. 2^2 = 4 , 4^4 = 16, etc.
if you're looking for more numbers, three more are: 12^12 = 144, 14^14 = 196, 16^16 =256

Marina CMI [18]3 years ago

5 0

This is a quadratic sequence.
16-4=12 and 36-16=20
20-12=8
there is a common difference of 8, so you add 12 to the first term to get the second term, that means you will add 8+12=20 to the second term to get the third term, then you will add 8+20=28 to the third term to get the fourth term, and so on... it's called a common difference.
so
2nd term: 4+12=16
3rd term: 16+(8+12)=16+(20)=36
4th term: 36+(8+20)=36+(28)=64
5th term: 64+(8+28)=64+(36)=100

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Convert the fraction to decimal and decimal into the fraction 10.222

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3 years ago

Read 2 more answers

Can someone tell me step by step how to solve this?

sasho [114]

Answer:

Step-by-step explanation:

I take it that this is some sort of ratio. Start by solving the brackets on the left.

Brackets: 1/3 - 1/10

Brackets: 10/30 - 3/30

Brackets: 7/30

Brackets^2: 49/900

Brackets^2 - 1/5: 49/900 - 180/900

Brackets^2 - 1/5: -131/900

(2/5)^2 = 4/25

The way this read, it should be

$\frac{\frac{-131}{900} }{\frac{4}{25} }$

Which when you invert and multiply becomes

$\frac{-131}{900} * \frac{25}{4}$

which finally becomes

$\frac{-131}{144}$

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3 years ago

A random sample of n = 45 observations from a quantitative population produced a mean x = 2.5 and a standard deviation s = 0.26.

oee [108]

Answer:

P-value (t=2.58) = 0.0066.

Note: as we are using the sample standard deviation, a t-statistic is appropiate instead os a z-statistic.

As the P-value (0.0066) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the population mean μ exceeds 2.4.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the population mean μ exceeds 2.4.

Then, the null and alternative hypothesis are:

$H_0: \mu=2.4\\\\H_a:\mu> 2.4$

The significance level is 0.05.

The sample has a size n=45.

The sample mean is M=2.5.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=0.26.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{0.26}{\sqrt{45}}=0.0388$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{2.5-2.4}{0.0388}=\dfrac{0.1}{0.0388}=2.58$

The degrees of freedom for this sample size are:

$df=n-1=45-1=44$

This test is a right-tailed test, with 44 degrees of freedom and t=2.58, so the P-value for this test is calculated as (using a t-table):

$P-value=P(t>2.5801)=0.0066$

As the P-value (0.0066) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the population mean μ exceeds 2.4.

7 0

3 years ago