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Zigmanuir [339]
3 years ago
10

Mario's class voted on a location for a field trip. 3/4 of the class voted for the museum. 1/8 of the class voted for the zoo. T

he rest of the class voted for the nature park. What fraction of the class voted for the nature park?
Mathematics
1 answer:
elixir [45]3 years ago
8 0
1 - 3/4 - 1/8 = 8/8 -6/8 -1/8 = (8 -6 -1)/8 = 1/8

1/8 of the class voted for the nature park, assuming the whole class voted once only.
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PLEASE HELP!!!!! 3, 8, 13, 18, 23, ....<br><br> The recursive formula for this sequence is:
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Answer:

a₈ = 37

Step-by-step explanation:

The given arithmetic sequence is: 3, 8, 13, 18, 23, . . .

The recursive formula for the sequence is: $ a_n = a_{n - 1} + 5 $

Here, $ a_n $ represents the $ n^{th} $ of the sequence.

And, $ a_{n - 1} $ represents the $ (n - 1)^{th} $ of the sequence.

'+5' denotes that '5' is added to the $ (n - 1)^{th} $ term to get the $ n^{th} $ term. In other words, the difference between two consecutive numbers in the sequence is 5.

Now, we are asked to find a₈ i.e., n =8.

Substituting in the recursive formula we get: a₈ = a₍₈₋ ₁₎ + 5 = a₇ + 5.

So, to determine a₈ we need to know a₇. From the sequence we see that a₅ = 23.

⇒ a₆ = 23 + 5 = 28.

⇒ a₇ = 28 + 5 = 32.

⇒ a₈ = 32 + 5 = 37.

Therefore, the $ 8^{th} $ term of the sequence is 37.

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2 years ago
owl monkey sleep during the day waking about 15minutes after sundown to find food.At midnight,they rest for an hour or two,then
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Answer:

x\leq 9

Step-by-step explanation:

Subtract 7 From both sides

x+7-7\le \:16-7

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10 cm<br> 8 cm<br> 2.5 cm<br> 8 cm<br> 6 cm<br> 2.5 cm <br> what’s the surface area
DIA [1.3K]

Answer:

10+8+2.5+8+6+2.5=37 .

ok you are understand or bot

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3 years ago
The local oil changing business is very busy on Saturday mornings and is considering expanding. A national study of similar busi
eimsori [14]

Answer:

t=\frac{4.2-3.6}{\frac{1.4}{\sqrt{16}}}=1.714

Reject the null hypothesis if the observed "t" value is less than -2.131 or higher than 2.131  

Rejection Zone: t_{calculated} or t_{calculated}>2.131

In our case since our calculated value is not on the rejection zone we don't have enough evidence to reject the null hypothesis at 5% of significance.

Step-by-step explanation:

Previous concepts  and data given  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

\bar X=4.2 represent the sample mean  

s=1.4 represent the sample standard deviation  

n=16 represent the sample selected  

\alpha=0.05 significance level  

State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to check if the mean is different from 3.6, the system of hypothesis would be:    

Null hypothesis:\mu = 3.6    

Alternative hypothesis:\mu \neq 3.6    

If we analyze the size for the sample is < 30 and we know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:    

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)    

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".    

Calculate the statistic  

We can replace in formula (1) the info given like this:    

t=\frac{4.2-3.6}{\frac{1.4}{\sqrt{16}}}=1.714

Critical values

On this case since we have a bilateral test we need to critical values. We need to use the t distribution with df=n-1=16-1=15 degrees of freedom. The value for \alpha=0.05 and \alpha/2=0.025 so we need to find on the t distribution with 15 degrees of freedom two values that accumulates 0.025 of the ara on each tail. We can use the following excel codes:

"=T.INV(0.025,15)" "=T.INV(1-0.025,15)"

And we got t_{crit}=\pm 2.131    

So the decision on this case would be:

Reject the null hypothesis if the observed "t" value is less than -2.131 or higher than 2.131  

Rejection Zone: t_{calculated} or t_{calculated}>2.131

Conclusion    

In our case since our calculated value is not on the rejection zone we don't have enough evidence to reject the null hypothesis at 5% of significance.

3 0
3 years ago
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