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Feliz [49]
2 years ago
8

If the​ principal, interest​ rate, or time in a simple interest problem is​ doubled, and the other two quantities remain​ consta

nt, how does the simple interest amount​ change? Explain.
Choose the correct answer below.
A.
The simple interest amount is found by multiplying the principal and interest rate and dividing by time.​ So, if the principal or interest rate is​ doubled, the interest amount will be doubled. If the time is​ doubled, the interest amount will be halved.
B.
The simple interest amount is found by multiplying the principal and time and dividing by the interest rate.​ So, if the principal or time is​ doubled, the interest amount will be doubled. If the interest rate is​ doubled, the interest amount will be halved.
C.
The simple interest amount is found by multiplying the​ principal, interest​ rate, and time.​ So, if any one of these values is​ doubled, it will cause the interest amount to be doubled.
D.
The simple interest amount is found by multiplying the​ principal, interest​ rate, and time.​ So, if any one of these values is​ doubled, it will cause the interest amount to be quadrupled.
Mathematics
1 answer:
VashaNatasha [74]2 years ago
5 0

In this question we have to find the effect of doubling principal, rate or time .

First we have to check the formula which is

I = P rt

As we see that interest is the product of principal, rate and time. So if any of these three doubles, that is if any of these three is twice of the original value, the interest gets doubled.

SO the correct option is C.

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A company employs two shifts of workers. Each shift produces a type of gasket where the thickness is the critical dimension. The
AURORKA [14]

Answer:

(−0.103371 ; 0.063371) ;

No ;

( -0.0463642, 0.0063642)

Step-by-step explanation:

Shift 1:

Sample size, n1 = 30

Mean, m1 = 10.53 mm ; Standard deviation, s1 = 0.14mm

Shift 2:

Sample size, n2 = 25

Mean, m2 = 10.55 ; Standard deviation, s2 = 0.17

Mean difference ; μ1 - μ2

Zcritical at 95% confidence interval = 1.96

Using the relation :

(m1 - m2) ± Zcritical * (s1²/n1 + s2²/n2)

(10.53-10.55) ± 1.96*sqrt(0.14^2/30 + 0.17^2/25)

Lower boundary :

-0.02 - 0.0833710 = −0.103371

Upper boundary :

-0.02 + 0.0833710 = 0.063371

(−0.103371 ; 0.063371)

B.)

We cannot conclude that gasket from shift 2 are on average wider Than gasket from shift 1, since the interval contains 0.

C.)

For sample size :

n1 = 300 ; n2 = 250

(10.53-10.55) ± 1.96*sqrt(0.14^2/300 + 0.17^2/250)

Lower boundary :

-0.02 - 0.0263642 = −0.0463642

Upper boundary :

-0.02 + 0.0263642 = 0.0063642

( -0.0463642, 0.0063642)

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How does the value of log Subscript 2 Baseline 100 compare with the value of Log Subscript 6 Baseline 20?.
attashe74 [19]

You can use the change of base formula to get

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and also

\log_6(20) = \frac{\log(20)}{\log(6)} \approx 1.671950

In general, the change of base formula is

\log_b(x) = \frac{\log(x)}{\log(b)}

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