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2 years ago

If you put $750 into a savings account that earns 6.2%, how much interest will you receive at the end of 7 years?

Mathematics

1 answer:

frozen [14]2 years ago

5 0

Answer: You will receive $ 325.5 as interest at the end of 7 years.

Step-by-step explanation:

Given: Principal amount = $750

Rate of interest = 6.2% = 0.062

Time = 7 years

Interest = Principal x Rate x Time

Inetrest = 750 x 0.062 x 7

= $ 325.5

Hence, you will receive $ 325.5 as interest at the end of 7 years.

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I need help with this problemmm, someone help mee!!.

Varvara68 [4.7K]

Answer:

∠A=30°

Step-by-step explanation:

let's call the angle at the bottom B and the one next to the 130 C

∠A=3x-6

∠C=180-130= 50°

vertical angles are equal∠B =8x+4

∠A+∠C+∠B=180°

find x:

3x-6+50+8x+4= 180°

11x+48=180°

11x= 132

x=12

find ∠A

∠A=3x-6= 3(12)-6 =30°

5 0

3 years ago

Tell me the answer and help me explain please

DENIUS [597]

Answer:

The answer is 9/10

Step-by-step explanation:

8/10 is the same thing as ⅘, ⅘ is just simplified

⅖ is less than ⅘ so it would come before c

6/5 is more than 1 so it wouldn’t be on that number line.

6 0

3 years ago

Read 2 more answers

The mean student loan debt for college graduates in Illinois is $30000 with a standard deviation of $9000. Suppose a random samp

Nataly [62]

Answer:

the probability that the mean student loan debt for these people is between $31000 and $33000 is 0.1331

Step-by-step explanation:

Given that:

Mean = 30000

Standard deviation = 9000

sample size = 100

The probability that the mean student loan debt for these people is between $31000 and $33000 can be computed as:

$P(31000 < X < 33000) = P( X \leq 33000) - P (X \leq 31000)$

$P(31000 < X < 33000) = P( \dfrac{X - 30000}{\dfrac{\sigma}{\sqrt{n}}} \leq \dfrac{33000 - 30000}{\dfrac{9000}{\sqrt{100}}} )- P( \dfrac{X - 30000}{\dfrac{\sigma}{\sqrt{n}}} \leq \dfrac{31000 - 30000}{\dfrac{9000}{\sqrt{100}}} )$

$P(31000 < X < 33000) = P( Z \leq \dfrac{33000 - 30000}{\dfrac{9000}{\sqrt{100}}} )- P(Z \leq \dfrac{31000 - 30000}{\dfrac{9000}{\sqrt{100}}} )$

$P(31000 < X < 33000) = P( Z \leq \dfrac{3000}{\dfrac{9000}{10}}}) -P(Z \leq \dfrac{1000}{\dfrac{9000}{10}}})$

$P(31000 < X < 33000) = P( Z \leq 3.33)-P(Z \leq 1.11})$

From Z tables:

$P(31000 < X$

Therefore; the probability that the mean student loan debt for these people is between $31000 and $33000 is 0.1331

8 0

3 years ago

A tennis tournament has 2n contestants. We want to pair them up for the first round of singles matches. Show that the number of

ddd [48]

There are

$\dbinom{2n}2 = \dfrac{(2n)!}{2! (2n-2)!}$

ways of pairing up any 2 members from the pool of $2n$ contestants. Note that

$(2n)! = 1\times2\times3\times4\times\cdots\times(2n-2)\times(2n-1)\times(2n) = (2n-2)! \times(2n-1) \times(2n)$

so that

$\dbinom{2n}2 = \dfrac{(2n)\times(2n-1)\times(2n-2)!}{2! (2n-2)!} = \boxed{n(2n-1)}$

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2 years ago

Given the functions, and g(x) = 3x 2 + 2, perform the indicated operation. f(g(x))

DIA [1.3K]

I’m not quite sure if I did this right,but 6x+3h

8 0

3 years ago

Read 2 more answers