Express each rate as a unit rate $46 for 5 toys

Find the sum of the first 6 terms of the sequence: –3, 1, 5, 9, . . .A) 45B) 42C) 48D) 25

Leni [432]

SOLUTION

From the sequence give

–3, 1, 5, 9, . . .

The first term, a = -3

The common difference, d = 4 (gotten by adding 4 to the next term).

The number of terms required n = 6.

Formula for sum of an arithmetic sequence is given by

$S_n=\frac{n}{2}\lbrack2a+(n-1)d\rbrack$

Substituting these values into the equation above we have

$\begin{gathered} S_n=\frac{n}{2}\lbrack2a+(n-1)d\rbrack \\ S_6=\frac{6}{2}\lbrack2\times-3+(6-1)4\rbrack \\ S_6=3\lbrack-6+(5)4\rbrack \\ S_6=3\lbrack-6+20\rbrack \\ S_6=3\lbrack14\rbrack \\ S_6=42 \end{gathered}$

Hence, the answer is 42, option B

3 0

8 months ago

A tire manufacturer is considering a newly designed tread pattern for its all-weather tires. Tests have indicated that these tir

Nataly_w [17]

Answer:

With outlier (102)

$t=\frac{126.2-125}{\frac{9.138}{\sqrt{10}}}=0.415$

$p_v =P(t_{9}>0.415)=0.344$

If we compare the p value and a significance level assumed $\alpha=0.05$ 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 true mean is not significantly higher than 125 feet at 5% of significance.

Without outlier (102)

$t=\frac{128.89-125}{\frac{3.551}{\sqrt{9}}}=3.285$

$p_v =P(t_{8}>3.285)=0.0056$

And we conclude that we reject the null hypothesis since $p_v$ . So the final conclusion would be not use the method since the value of 102 observed can be a potential outlier, removing this value we see that we reject the null hypothesis and we have a significant result that the true mean is higher than 125.

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".

Data: 129, 128, 130, 132, 135, 123, 102, 125, 128, 130

We can calculate the mean with the following formulas:

$\bar X =\frac{\sum_{i=1}^n X_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$\bar X=126.2$ represent the sample mean

$s=9.138$ represent the sample standard deviation

n=10 represent the sample selected

$\alpha$ significance level

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is significantly higher than 125, the system of hypothesis would be:

Null hypothesis: $\mu \leq 125$

Alternative hypothesis: $\mu > 125$

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:

$t=\frac{126.2-125}{\frac{9.138}{\sqrt{10}}}=0.415$

P-value

First we need to calculate the degrees of freedom given by:

$df=n-1=10-1= 9$

Then since is a right tailed sided test the p value would be:

$p_v =P(t_{9}>0.415)=0.344$

Conclusion

If we compare the p value and a significance level assumed $\alpha=0.05$ 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 true mean is not significantly higher than 125 feet at 5% of significance.

Using all the data we see that we don;t have enough info to conclude that the true mean is higher than 125, but if we see careful 9 of the 10 values are over the limit of 125 feet. And if we repeat the procedure with the outlier of 102. We got this:

$\bar X=128.89$ represent the sample mean

$s=3.551$ represent the sample standard deviation

$t=\frac{128.89-125}{\frac{3.551}{\sqrt{9}}}=3.285$

First we need to calculate the degrees of freedom given by:

$df=n-1=9-1= 8$

Then since is a right tailed sided test the p value would be:

$p_v =P(t_{8}>3.285)=0.0056$

And we conclude that we reject the null hypothesis since $p_v$ . So the final conclusion would be not use the method since the value of 102 observed can be a potential outlier removing this value we see that we reject the null hypothesis and we have a significant result that the true mean is higher than 125.

8 0

2 years ago

Beatrix ate 21 raisins. This represents 30% of the box of raisins. How many raisins were originally in the box?

leonid [27]

21/0.3= 70
there were 70 raisins in the box

8 0

3 years ago

Among fatal plane crashes that occurred during the past 5050 years, 699699 were due to pilot error, 8888 were due to other hum

lapo4ka [179]

Answer: Crashes due to pilot error is the most serious threat to aviation safety as it has highest relative frequency.

Step-by-step explanation:

Since we have given that

Number of crashes due to pilot error = 699

Number of crashes due to other human error = 88

Number of crashes due to mechanical problems = 156

Number of crashes due to sabotage = 164

Total number of crashes = 699+88+156+164 = 1167

So, Relative frequency distribution :

Pilot error = $\dfrac{699}{1167}\times 100=59.89\%$

Human error = $\dfrac{88}{1167}\times 100=7.54\%$

Mechanical problems = $\dfrac{156}{1167}\times 100=13.36\%$

Sabtoge = $\dfrac{164}{1167}\times 100=14.05\%$

Crashes due to pilot error is the most serious threat to aviation safety as it has highest relative frequency.

6 0

3 years ago

A system of 2 linear equations is ___ dimensional. It is a graph of _____ lines. The solutions can be _____ solution if the grap

dexar [7]

Considering High School level question, answer can be written as:

A system of 2 linear equations is [two] dimensional. It is a graph of [two] lines. The solutions can be [unique] solution if the graph intersects. [No] solution if the lines are parallel - meaning they have the same slope, or [Infinitely many] solutions if they are the same line.

Explanation:

when two lines are drawn on a two-dimensional plane then there are only three possible cases:

Case1: lines will intersect

In that case you will get a unique solution at the intersection point.

Case2: lines are parallel but don't touch each other

In that case there will be no point which lies on both lines so No solution.

Case3: lines are overlapping.

In that case all the points lies on both lines so infinitely many solutions.

3 0

2 years ago