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Degger [83]
3 years ago
9

If the APY of a savings account is 3.2%, and if the principal in the savings account was $2400 for an entire year, what will the

balance of the savings account be after all the interest is paid for the year?
Mathematics
2 answers:
Diano4ka-milaya [45]3 years ago
8 0

Answer:

2476.80

Step-by-step explanation:

Setler79 [48]3 years ago
3 0
$2400.00+3.2%= $2476.80
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10 PTS!! (5 + 2/8) - (3 7/8)
Artyom0805 [142]
The answer should be 1 3/8
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3 years ago
Read 2 more answers
NEED HELP ASAP.....At the 2012 Summer Olympic Games in London, in the Men's Shot Put qualifying round, the distances ranged from
jeka57 [31]

Answer:

About 6 athletes.

Step-by-step explanation:

We are concerned only with the percent of athletes that are above one standard deviation.

Between one standard deviation and two standard deviations, there is 13.6% of the athletes.

From 2 to 3, there is 2.1%. Above 3, there is 0.2%.

The sum of those percentages is 15.9% of 40 = 0.159 · 40 = 6.36 or about 6 athletes.

Hope this helps.

6 0
4 years ago
Points F(3, 3), K(6, 5), J(8, 2) and L(5, 0) lie in a coordinate plane. Which lines are perpendicular?
Romashka-Z-Leto [24]
To determine if a line is perpendicular to another, you must first determine the slope...
m = y1-y2/x1-x2
m of FK = 3-5/3-6 = -2/-3 = 2/3
m of FJ = 3-2/3-8 = 1/-5
m of FL = 3-0/3-5 = 3/-2
m of KJ = 5-2/6-8 = 3/-2
m of KL = 5-0/6-5 = 5
m of JL = 2-0/ 8-5 = 2/3

In order for two lines to be perpendicular, their slopes must be opposite reciprocals...
FK is perpendicular to FL
FK is perpendicular to KJ
JL is perpendicular to FL
JL is perpendicular to KJ
FJ is perpendicular to KL
8 0
3 years ago
34​% of college students say they use credit cards because of the rewards program. You randomly select 10 college students and a
finlep [7]

Answer:

a) There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

b) There is a 71.62% probability that more than two students use credit cards because of the rewards program.

c) There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

Step-by-step explanation:

There are only two possible outcomes. Either the student use credit cards because of the rewards program, or they use for other reason. So, we can solve this problem by the binomial distribution.

Binomial probability

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

In which C_{n,x} is the number of different combinatios of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And \pi is the probability of X happening.

In this problem, we have that:

10 student are sampled, so n = 10

34% of college students say they use credit cards because of the rewards program, so \pi = 0.34

(a) exactly​ two

This is P(X = 2).

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873

There is a 18.73% probability that exactly two students use credit cards because of the rewards program.

(b) more than​ two

This is P(X > 2).

Either a value is larger than two, or it is smaller of equal. The sum of the decimal probabilities must be 1. So:

P(X \leq 2) + P(X > 2) = 1

P(X > 2) = 1 - P(X \leq 2)

In which

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)

So

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

P(X = 0) = C_{10,0}.(0.34)^{0}.(0.66)^{10} = 0.0157

P(X = 1) = C_{10,1}.(0.34)^{1}.(0.66)^{9} = 0.0808

P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0157 + 0.0808 + 0.1873 = 0.2838

P(X > 2) = 1 - P(X \leq 2) = 1 - 0.2838 = 0.7162

There is a 71.62% probability that more than two students use credit cards because of the rewards program.

(c) between two and five inclusive

This is:

P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)

P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}

P(X = 2) = C_{10,2}.(0.34)^{2}.(0.66)^{8} = 0.1873

P(X = 3) = C_{10,3}.(0.34)^{3}.(0.66)^{7} = 0.2573

P(X = 4) = C_{10,4}.(0.34)^{4}.(0.66)^{6} = 0.2320

P(X = 5) = C_{10,5}.(0.34)^{5}.(0.66)^{5} = 0.1434

P = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1873 + 0.2573 + 0.2320 + 0.1434 = 0.82

There is a 82% probability that between two and five students, inclusive, use credit cards because of the rewards program.

6 0
3 years ago
Pls explain how to solve this elimination problem!!
trapecia [35]
Multiply the second equation by -2. This will cancel out the y
6 0
3 years ago
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