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3 years ago

Hello is it anyone up here that can help me with my question please

Mathematics

1 answer:

jok3333 [9.3K]3 years ago

7 0

Simple interest is basically the cost of borrowing money over a period of time. So if you have borrowed $110.00 at 5% for two years, you will multiply the 5% by the two years (presuming that it is 5% annual percentage rate (APR). So, You will multiply the 110 by 10% (or .1) to get $11 dollars of simple interest.

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Find the slope of (-3,3) and (-21,18)

ZanzabumX [31]

Answer:

-0.8333

Step-by-step explanation:

used omni calculator

https://www.omnicalculator.com/math/slope

3 0

3 years ago

The amount of money spent on textbooks per year for students is approximately normal.

Ostrovityanka [42]

Answer:a

$336.04 < \mu < 443.96$

The margin of error will increase

The margin of error will decreases

The 99% confidence interval is $0.4107 < p < 0.4293$

Step-by-step explanation:

From the question we are told that

The sample size $n = 19$

The sample mean is $\= x = \$\ 390$

The standard deviation is $\sigma = \$ \ 120$

Given that the confidence level is 95% then the level of significance is mathematically represented as

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha = 0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table

$Z_{\frac{\alpha }{2} } = 1.96$

The margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma}{\sqrt{n} }$

=> $E = 1.96 * \frac{120}{\sqrt{19} }$

=> $E = 53.96$

The 95% confidence interval is

$\= x - E < \mu < \= x + E$

=> $390 - 53.96 < \mu < 390 - 53.96$

=> $336.04 < \mu < 443.96$

When the confidence level increases the $Z_{\frac{\alpha }{2} }$ also increases which increases the margin of error hence the confidence level becomes wider