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MrMuchimi
3 years ago
7

If the simple interest on ​$7,000 for 3 years is ​$1​890, then what is the interest​ rate?

Mathematics
1 answer:
katovenus [111]3 years ago
5 0

Answer:

I = Prt

Interest = Principal x rate x time

Step-by-step explanation:

1890 = 7000(3)r

1890 = 21000r

r = 1890/21000

r = .09 or 9%

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In a normal distribution, what is the probability that a randomly selected data value will fall within two standard deviations o
Zepler [3.9K]
The answer is 95%

This is something you should memorize. Specifically it is from the Empirical Rule (or 68-95-99.7 rule) which gives rough approximations of areas under the curve. 
5 0
3 years ago
How much simple interest is earned on $6000 principal, with an annual interest rate of 3.7% over 5 years?
diamong [38]
I = PRT
P (principal) = 6000
R (rate) = 3.7%.....u need decimal form = 0.037
T (time) = 5

I = (6000)(0.037)(5)
I = 1110.00 <===

u dont need years as a decimal...u need rate as a decimal
8 0
3 years ago
Read 2 more answers
50 Points:
kodGreya [7K]

Answer:

(-2,3)

Step-by-step explanation:

For the point to be 2/3 of the way; it means it divides AB into the ratio 2 to 1

Now, we can use the internal section formula to get the coordinates of this point

(x,y) = (nx1 + mx2)/(m + n), (ny1 + my2)/(m + n)

where (m,n) = (2,1)

(x1,y1) = (-4,-1)

(x2,y2) = (5,5)

(x,y) = (1(-4) + 2(-1)/(1+2), (1(-1)+2(5)/(1+2)

(x,y) = (-4-2)/3, (-1 + 10)/3

(x,y) = (-2,3)

4 0
2 years ago
Jaylen will deposit $1,750 in an account that earns 6% simple interest every year. His best friend, Eli, will deposit $1,600 in
Novay_Z [31]

Answer:

Jaylen- 2275

Eli- 2405.85

Jaylen

1750 x 0.06 x 5 =525

525 + 1750 = 2275

He makes 525 simple interest in those 5 years so you add that to the initial amount, to find his total after 5 years.

Eli

A= 1600(1+0.085)^5

1600(1.085)^5

2405.85

Hope this helps ^-^

6 0
2 years ago
Which of the following is the expansion of (3c + d2)6?
vovangra [49]

Answer: Option A) 729c^6 + 1,458c^5d^2 + 1,215c^4d^4 + 540c^3d^6 + 135c^2d^8 + 18cd^{10} + d^{12} is the correct expansion.

Explanation:

on applying binomial theorem,  (a+b)^n=\sum_{r=0}^{n} \frac{n!}{r!(n-r)!} a^{n-r} b^r

Here a=3c, b=d^2 and n=6,

Thus, (3c+d^2)^6=\sum_{r=0}^{6} \frac{6!}{r!(6-r)!} (3c)^{n-r} (d^2)^r

⇒ (3c+d^2)^6= \frac{6!}{(6-0)!0!} (3c)^{6-0}.(d^2)^0+\frac{6!}{(6-1)!1!} (3c)^{6-1}.(d^2)^1+\frac{6!}{(6-2)!2!} (3c)^{6-2}.(d^2)^2+\frac{6!}{(6-3)!3!} (3c)^{6-3}.(d^2)^3+\frac{6!}{(6-4)!4!} (3c)^{6-4}.(d^2)^4+\frac{6!}{(6-5)!5!} (3c)^{6-5}.(d^2)^5+\frac{6!}{(6-6)!6!} (3c)^{6-6}.(d^2)^6

⇒(3c+d^2)^6= \frac{6!}{(6-)!0!} (3c)^6.d^0+\frac{6!}{(5)!1!} (3c)^5.d^2+\frac{6!}{(4)!2!} (3c)^4.d^4+\frac{6!}{(6-3)!3!} (3c)^3.d^6+\frac{6!}{(2)!4!} (3c)^2.d^8+\frac{6!}{(1)!5!} (3c).d^{10}+\frac{6!}{(0)!6!} (3c)^0.d^{12}

⇒(3c+d^2)^6=(3c)^6.d^0+\frac{720}{120} (3c)^5.d^2+\frac{720}{48} (3c)^4.d^4+\frac{720}{36} (3c)^3.d^6+\frac{720}{48} (3c)^2.d^8+\frac{720}{120} (3c).d^{10}+.d^{12}

⇒(3c+d^2)^6=729c^6 + 1,458c^5d^2 + 1,215c^4d^4 + 540c^3d^6 + 135c^2d^8 + 18cd^{10} + d^{12}

3 0
3 years ago
Read 2 more answers
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