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3 years ago

11

The difference between the annual and semi-annual compound interest on a sum of money is Rs 482 at the rate of 20 % per annum fo

r 2 years. Find the sum.

1 answer:

Effectus [21]3 years ago

6 0

Just multiply 482 by .20 then times by 2

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What value of c makes the statement true?

Vanyuwa [196]

$c-(-17.6)=-4 \\\\ c+17.6=-4 \\\\ c=-4-17.6 \\\\ \boxed{c=-21.6\to \ variant \ A}$

6 0

3 years ago

Read 2 more answers

Which is a step in the process of calculating successive discounts of 8% and 10% on a $50 item

Paraphin [41]

Answer:

The process of calculating successive discounts of 8% and 10% on a $50 item is take 10% of $46.

Step-by-step explanation:

As given

successive discounts of 8% and 10% on a $50 item .

First find out for 8 % discount

8% is written in the decimal form

= 0.08

8 % of $50 item = 0.08 × 50

= $ 4

Price of item after 8% discount = 50 - 4

= $46

First find out for 10 % discount

10% is written in the decimal form

= 0.1

8 % of $48 item = 0.1× 46

= $4.6

Price of item after 8% discount = 46 - 4.6

= $41.4

Therefore in the successive discounts of 8% and 10% on a $50 item is $41.4 .

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

How would you write this in simplest form 4p-5 (p+6)=

pychu [463]

Answer:

the answer is -p-30

Step-by-step explanation:

first you would distribute -5 to p and 6 = 4p-5p-30

second combine the like terms = -p-30

lastly it cannot be factored down any farther so the answer is just -p-30

7 0

3 years ago

Austin borrowed money from a credit union for 4 years and was charged simple interest at an annual rate of 5% the total interest

Reika [66]

Given:

Rate of simple interest = 5%

Time = 4 years

Total interest = $160

To find:

The amount borrowed by Austin from a credit union.

Solution:

The formula for simple interest is:

$I=\dfrac{P\times r\times t}{100}$

Where, P is principal, r is the rate of interest and t is the number of years.

Putting $I=160,r=5,t=4$ in the above formula, we get

$160=\dfrac{P\times 5\times 4}{100}$

$160=\dfrac{20P}{100}$

$160=\dfrac{P}{5}$

Multiply both sides by 5.

$800=P$

Therefore, Austin borrowed $800 from a credit union for 4 years.

8 0

3 years ago