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3 years ago

How do you calculate the balance in an account, with interest?

Mathematics

1 answer:

pentagon [3]3 years ago

5 0

Answer:

Divide your interest rate by the number of payments you'll make in the year (interest rates are expressed annually). ...

Multiply it by the balance of your loan, which for the first payment, will be your whole principal amount.

Step-by-step explanation:

The ending balance, or future value, of an account with simple interest can be calculated using the following formula: Using the prior example of a $1000 account with a 10% rate, after 3 years the balance would be $1300. This can be determined by multiplying the $1000 original balance times [1+(10%)(3)], or times 1.30.

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For an essay contest, Annika needs to write 400 words with an absolute deviation of less than 10 words. Using w as a variable, w

saw5 [17]

Answer:

The absolute value inequality to represent the situation is presented as follows;

$\left | w -400 \right |$

Step-by-step explanation:

The given information are;

The number of words Annika needs to write = 400 words

The absolute deviation of the number of words Annika needs to write = 10 words

When we let the number of words Annika write = w, we have;

The absolute value inequality given that the absolute deviation of the number of words is 10 is given as follows;

$\left | w -400 \right |$

Therefore we have;

$\left | w -400 \right |$ gives the statement, -10 < w - 400 < 10.

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3 years ago

Simplify the square root of 27 over 12

Hunter-Best [27]

$\sqrt{ \frac{27}{12} }$

${ \frac{ \sqrt{27} {} }{ \sqrt{12}}$

$\frac{ \sqrt{9 * 3} }{ \sqrt{4 * 3} }$

$\frac{ \sqrt{9} \sqrt{3} }{ \sqrt{4} \sqrt{3} }$

$\frac{3 \sqrt{3} }{2 \sqrt{3} }$

$\frac{3}{2}$

$1 \frac{1}{2}$

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4 years ago

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Inessa [10]

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3 years ago

Least common multiple. of 5 ,10 and 16

Andreyy89

Answer:

Step-by-step explanation:

There is no lcm other than 1 between these 3 numbers

5 factors: 1, 5

10 factors: 10,1,5,2

16 factors:16,1,4,4,8,2

All of these have the number one in common, hence the answer

4 0

3 years ago

Read 2 more answers

Construct a 90% confidence interval for μ1-μ2 with the sample statistics for mean calorie content of two bakeries' specialty pi

DIA [1.3K]

Answer:

The 90% confidence interval for the difference in mean (μ₁ - μ₂) for the two bakeries is; (49) < μ₁ - μ₂ < (289)

Step-by-step explanation:

The given data are;

Bakery A

$\overline x_1$ = 1,880 cal

s₁ = 148 cal

n₁ = 10

Bakery B

$\overline x_2$ = 1,711 cal

s₂ = 192 cal

n₂ = 10

$\left (\bar{x}_1-\bar{x}_{2} \right ) - t_{c}\cdot \hat \sigma \sqrt{\dfrac{1}{n_{1}}+\dfrac{1}{n_{2}}}< \mu _{1}-\mu _{2}< \left (\bar{x}_1-\bar{x}_{2} \right ) + t_{c}\cdot \hat \sigma \sqrt{\dfrac{1}{n_{1}}+\dfrac{1}{n_{2}}}$

df = n₁ + n₂ - 2

∴ df = 10 + 18 - 2 = 26

From the t-table, we have, for two tails, $t_c$ = 1.706

$\hat{\sigma} =\sqrt{\dfrac{\left ( n_{1}-1 \right )\cdot s_{1}^{2} +\left ( n_{2}-1 \right )\cdot s_{2}^{2}}{n_{1}+n_{2}-2}}$

$\hat{\sigma} =\sqrt{\dfrac{\left ( 10-1 \right )\cdot 148^{2} +\left ( 18-1 \right )\cdot 192^{2}}{10+18-2}}= 178.004321469$

$\hat \sigma$ ≈ 178

Therefore, we get;

$\left (1,880-1,711 \right ) - 1.706\times178 \sqrt{\dfrac{1}{10}+\dfrac{1}{18}}< \mu _{1}-\mu _{2}< \left (1,880-1,711 \right ) + 1.706\times178 \sqrt{\dfrac{1}{10}+\dfrac{1}{18}}$

Which gives;

$169 - \dfrac{75917\cdot \sqrt{35} }{3,750} < \mu _{1}-\mu _{2}< 169 + \dfrac{75917\cdot \sqrt{35} }{3,750}$

Therefore, by rounding to the nearest integer, we have;

The 90% C.I. ≈ 49 < μ₁ - μ₂ < 289

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3 years ago