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wolverine [178]
3 years ago
6

Write an explicit formula for the following sequences. -4, -6, -8, -10...

Mathematics
1 answer:
LekaFEV [45]3 years ago
6 0

Answer:

tn=-2n-2

hope this helps :)

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Solve for x. x2 + x - 6 = 0 A. -2, -3 B. -2, 3 C. 2, -3 D. 2, 3
stiv31 [10]

Factor x^2 + x - 6

(x - 2)(x + 3) = 0

Solve for x

Ask yourself; When will (x - 2)(x + 3) equal zero?

When x - 2 = 0 or x + 3 = 0

Then solve for each of the two equations above

<em>x = 2, -3</em>

<u>Answer: C. 2, -3</u>

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In how many ways can 7 people line if A needs to be first or last and B needs to be in the middle
12345 [234]

Answer:

<em>Assuming person A and person B are part of the 7 people</em>, there are 10 ways the 7 people can line up if A needs to be first or last and B needs to be in the middle.

Step-by-step explanation:

There are 2 places for A (first, last) and 5 places for B (anywhere but first or last). 2 x 5 = 10.

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3 years ago
1.5x+0.75y what does x variable
Illusion [34]

Answer:

1.5

Step-by-step explanation:


3 0
3 years ago
3
siniylev [52]

Answer:

179

Step-by-step explanation:

The total number of animals in the river was 632.

A herd of 187 elephants and 266 zebras left the river area.

Adding them up, the total number of animals that left the river area is:

187 + 266 = 453

Therefore, the total number of animals left at the river area is the subtraction for the animals that left the river area from the number of animals that were there initially:

632 - 453 = 179

3 0
3 years ago
We are conducting a hypothesis test to determine if fewer than 80% of ST 311 Hybrid students complete all modules before class.
Juli2301 [7.4K]

Answer:

z=\frac{0.881 -0.8}{\sqrt{\frac{0.8(1-0.8)}{110}}}=2.124  

Null hypothesis:p\geq 0.8  

Alternative hypothesis:p < 0.8  

Since is a left tailed test the p value would be:  

p_v =P(Z  

So the p value obtained was a very high value and using the significance level assumed \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

Be Careful with the system of hypothesis!

If we conduct the test with the following hypothesis:

Null hypothesis:p\leq 0.8  

Alternative hypothesis:p > 0.8

p_v =P(Z>2.124)=0.013  

So the p value obtained was a very low value and using the significance level assumed \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis.

Step-by-step explanation:

1) Data given and notation  

n=110 represent the random sample taken

X=97 represent the students who completed the modules before class

\hat p=\frac{97}{110}=0.882 estimated proportion of students who completed the modules before class

p_o=0.8 is the value that we want to test

\alpha represent the significance level

z would represent the statistic (variable of interest)

p_v{/tex} represent the p value (variable of interest)  2) Concepts and formulas to use  We need to conduct a hypothesis in order to test the claim that the true proportion is less than 0.8 or 80%:  Null hypothesis:[tex]p\geq 0.8  

Alternative hypothesis:p < 0.8  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.881 -0.8}{\sqrt{\frac{0.8(1-0.8)}{110}}}=2.124  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level assumed \alpha=0.05. The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

p_v =P(Z  

So the p value obtained was a very high value and using the significance level assumed \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

Be Careful with the system of hypothesis!

If we conduct the test with the following hypothesis:

Null hypothesis:p\leq 0.8  

Alternative hypothesis:p > 0.8

p_v =P(Z>2.124)=0.013  

So the p value obtained was a very low value and using the significance level assumed \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis.

5 0
4 years ago
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