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liq [111]
3 years ago
12

A sum of rs 400 is lent for 3 yrs at the rate of 6% per annum. Find the interest.

Mathematics
1 answer:
Nimfa-mama [501]3 years ago
3 0
Interest is equal to rs 72

Simple Interest is computed by multiplying the principal to the interest rate and the time.

I = P * r * t

Principal = rs 400
Interest rate = 6%
Term / time = 3 yrs

I = 400 * 6% * 3
I = 72   the total interest that must be paid in 3 yrs.
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First you have to find out how many guys you have-1.) 32-9=23, then you divide the number of men by the number of students-2.) 23÷32=0.718, so the answer is 0.7
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Jonathan makes a weekly allowance of $25. He also makes $9.50 an hour at his job. Because of his age, he can work no more than 2
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Elliot has been running a lawn care business since 2000. He cuts grass, trims, and weed whacks yards for his customers throughou
pashok25 [27]
A.) Rate of change = (1350 - 750) / (2010 - 2000) = 600/10 = $60
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3 years ago
Here are two datasets:
fiasKO [112]

Answer:

Option a) Mean

Mean is affected a lot by the change in the last observation as the median remains the same.

Step-by-step explanation:

we are given the following in the question:

Data set A: 64, 65, 66, 68, 70, 71, 72

Data set B: 64, 65, 66, 68, 70, 71, 720

For data set A, the mean and median are 68.

For data set B:

Formula:

Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}

Mean =\displaystyle\frac{1124}{7} = 160.57

Median:\\\text{If n is odd, then}\\\\Median = \displaystyle\frac{n+1}{2}th ~term \\\\\text{If n is even, then}\\\\Median = \displaystyle\frac{\frac{n}{2}th~term + (\frac{n}{2}+1)th~term}{2}

Sorted data:

64, 65, 66, 68, 70, 71, 720

\text{Median} = \dfrac{7+1}{2}^{th} \text{ term} = 4^th\text{ term} = 68

Clearly, 720 is the is a outlier.

As seen mean is affected a lot by the change in the last observation as the median remains the same.

5 0
3 years ago
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
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