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4 years ago

The price of a jeans was reduced $6 per week for 7 weeks. By how much did the price of the jeans change over the 7 weeks?

Mathematics

2 answers:

ad-work [718]4 years ago

8 0

6•7=42, so the price changed $42 in 7 weeks.

kompoz [17]4 years ago

4 0

It was reduced by $42.00 by the end of the seven weeks.

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Four students are working in a group and are asked to find the length of BC. Which two students set up the problem correctly? Mo

Alika [10]

The best and the most correct answer among the choices provided by the question is the first choice. The two students that set up the problem correctly is <span> Moe: x = (5 - 1)2 + (1 - 4)2 and Jimmy: x2 = 32 + 42.</span> I hope my answer has come to your help. God bless and have a nice day ahead!

7 0

4 years ago

Read 2 more answers

Three assembly lines are used to produce a certain component for an airliner. To examine the production rate, a random

Katyanochek1 [597]

Answer:

a) Null hypothesis: $\mu_A =\mu_B =\mu C$

Alternative hypothesis: $\mu_i \neq \mu_j, i,j=A,B,C$

$SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2 =20.5$

$SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2 =12.333$

$SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2 =8.16667$

And we have this property

$SST=SS_{between}+SS_{within}$

The degrees of freedom for the numerator on this case is given by $df_{num}=df_{within}=k-1=3-1=2$ where k =3 represent the number of groups.

The degrees of freedom for the denominator on this case is given by $df_{den}=df_{between}=N-K=3*6-3=15$ .

And the total degrees of freedom would be $df=N-1=3*6 -1 =15$

The mean squares between groups are given by:

$MS_{between}= \frac{SS_{between}}{k-1}= \frac{12.333}{2}=6.166$

And the mean squares within are:

$MS_{within}= \frac{SS_{within}}{N-k}= \frac{8.1667}{15}=0.544$

And the F statistic is given by:

$F = \frac{MS_{betw}}{MS_{with}}= \frac{6.166}{0.544}= 11.326$

And the p value is given by:

$p_v= P(F_{2,15} >11.326) = 0.00105$

So then since the p value is lower then the significance level we have enough evidence to reject the null hypothesis and we conclude that we have at least on mean different between the 3 groups.

b) $(\bar X_B -\bar X_C) \pm t_{\alpha/2} \sqrt{\frac{s^2_B}{n_B} +\frac{s^2_C}{n_C}}$

The degrees of freedom are given by:

$df = n_B +n_C -2= 6+6-2=10$

The confidence level is 99% so then $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$ and the critical value would be: $t_{\alpha/2}=3.169$

The confidence interval would be given by:

$(43.333 -41.5) - 3.169 \sqrt{\frac{0.6667}{6} +\frac{0.7}{6}}= 0.321$

$(43.333 -41.5) + 3.169 \sqrt{\frac{0.6667}{6} +\frac{0.7}{6}}=3.345$

Step-by-step explanation:

Previous concepts

Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".

The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"

Part a

Null hypothesis: $\mu_A =\mu_B =\mu C$

Alternative hypothesis: $\mu_i \neq \mu_j, i,j=A,B,C$

If we assume that we have $3$ groups and on each group from $j=1,\dots,6$ we have $6$ individuals on each group we can define the following formulas of variation:

$SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2 =20.5$

$SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2 =12.333$

$SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2 =8.16667$

And we have this property

$SST=SS_{between}+SS_{within}$

The degrees of freedom for the numerator on this case is given by $df_{num}=df_{within}=k-1=3-1=2$ where k =3 represent the number of groups.

The degrees of freedom for the denominator on this case is given by $df_{den}=df_{between}=N-K=3*6-3=15$ .

And the total degrees of freedom would be $df=N-1=3*6 -1 =15$

The mean squares between groups are given by:

$MS_{between}= \frac{SS_{between}}{k-1}= \frac{12.333}{2}=6.166$

And the mean squares within are:

$MS_{within}= \frac{SS_{within}}{N-k}= \frac{8.1667}{15}=0.544$

And the F statistic is given by:

$F = \frac{MS_{betw}}{MS_{with}}= \frac{6.166}{0.544}= 11.326$

And the p value is given by:

$p_v= P(F_{2,15} >11.326) = 0.00105$

So then since the p value is lower then the significance level we have enough evidence to reject the null hypothesis and we conclude that we have at least on mean different between the 3 groups.

Part b

For this case the confidence interval for the difference woud be given by:

$(\bar X_B -\bar X_C) \pm t_{\alpha/2} \sqrt{\frac{s^2_B}{n_B} +\frac{s^2_C}{n_C}}$

The degrees of freedom are given by:

$df = n_B +n_C -2= 6+6-2=10$

The confidence level is 99% so then $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$ and the critical value would be: $t_{\alpha/2}=3.169$

The confidence interval would be given by:

$(43.333 -41.5) - 3.169 \sqrt{\frac{0.6667}{6} +\frac{0.7}{6}}= 0.321$

$(43.333 -41.5) + 3.169 \sqrt{\frac{0.6667}{6} +\frac{0.7}{6}}=3.345$

7 0

3 years ago

Selena drove at an average speed of 50.55 Miles per hours for 1.75 hours. She stopped at a rest stop and then drove at an averag

Likurg_2 [28]

The correct answer would be 13.2375.

8 0

3 years ago

Write a slope- intercept equation for a line with given characters m=-4/5, passes through (-4,-5)

aleksandr82 [10.1K]

$\bf (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-5})~\hspace{10em} \stackrel{slope}{m}\implies -\cfrac{4}{5} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-5)}=\stackrel{m}{-\cfrac{4}{5}}[x-\stackrel{x_1}{(-4)}]\implies y+5=-\cfrac{4}{5}(x+4) \\\\\\ y+5=-\cfrac{4}{5}x-\cfrac{16}{5}\implies y=-\cfrac{4}{5}x-\cfrac{16}{5}-5\implies y=-\cfrac{4}{5}x-\cfrac{41}{5}$

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3 years ago

independent variable. what is the independent variable? what are you deliberately choosing or changing?

liq [111]

The independent variable is the factor that you purposely change or control in order to see what effect it has. The variable that responds to the change in the independent variable is called the dependent variable. It depends on the independent variable. The independent variable is graphed on the x-axis.

To identify dependent studies variables, search for objects in your studies query or speculation that sees the result, effect, or outcome of changing the independent variable. The basic rule is to search for what reasons for reactions and what receives the results.

The statistical courting between variables is referred to as their correlation. A correlation might be high-quality, which means each variable circulates in the same direction, or bad, meaning that once one variable's price will increase, the opposite variables' values decrease.

Making sure that certain research variables are managed will increase the reliability and validity of the test, by ensuring that other causal consequences are removed. This safeguard makes it simpler for other researchers to repeat the experiment and comprehensively test the results.

Learn more about variables here brainly.com/question/2804470

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2 years ago