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3 years ago

Need help will give 14 points

Mathematics

1 answer:

azamat3 years ago

6 0

Answer:

1) False. 2) False.

Step-by-step explanation:

1) False. 2) False.

1) is false because it says, "A total of 40 students can type from 0-59 words per minute." 0-14w=5s, 15-29w=10s, 10-44w=14s, and 45-59w=20s. w= words per minutes s= number of students (BTW) 5+10+14+20=54 not 40

2) is false because it says, "The number of students who typed 45-59 words per minute is equal to the number of students who typed 15-39 and 0-14."

45-59w=20s

15-39w=10s and 0-14w=5s

10s+5s=15s

20s>15s

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J. Reexamine the sequence 20, 14, 8, 2, ... from the problem

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Answer:

The n th of the given sequence is $t_{n} = 26-6 n$

Step-by-step explanation:

<u>Step 1</u> :-

Given sequence is 20,14,8,2,.......this sequence in arithmetic progression but this sequence is decreasing sequence.

given first term is 20 and difference is $d = second term- first term = 14-20=-6$

now the nth term of given sequence is

by using formula $t_{n}=a+(n-1)d$

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final answer:-

$t_{n} = 26-6 n$

<u>verification</u>:-

$t_{n} = 26-6 n$

put n=1 we get first term is 20

put n=2 we get second term is 14

put n=3 we get third term is 8

put n=4 we get fourth term is 2

so the n th term of sequence is

$t_{n} = 26-6 n$

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3 years ago

Daily-high temperature measurements for 40 consecutive days are recorded for a particular city. The mean daily-high temperature

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Answer:

$t=\frac{21.5-22}{\frac{1.5}{\sqrt{40}}}=-2.108$

$p_v =P(t_{39}$

Conclusion

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the mean temperature actually its NOT significant less then 22 at 1% of signficance.

(D) P-val=0.021, fail to reject the null hypothesis

Step-by-step explanation:

1) Data given and notation

$\bar X=21.5$ represent the mean for the temperatures

$s=1.5$ represent the sample standard deviation

$n=40$ sample size

$\mu_o =22$ represent the value that we want to test

$\alpha=0.01$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is less than 22C, the system of hypothesis would be:

Null hypothesis: $\mu \geq 22$

Alternative hypothesis: $\mu < 22$

If we analyze the size for the sample is > 30 and we don't 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:

$t=\frac{21.5-22}{\frac{1.5}{\sqrt{40}}}=-2.108$

P-value

We can calculate the degrees of freedom like this:

$df=n-1=40-1=39$

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

$p_v =P(t_{39}$

Conclusion

The best option would be:

(D) P-val=0.021, fail to reject the null hypothesis

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