Write the equation (0,5) (2.0)

Sumara Kato's savings account has a balance of $1663. After 19 years, how much interest

lesantik [10]

Answer: $2936.27

Step-by-step explanation:

Present value = $1663

Rate = 5.5%

Time(n) = 19years

Future value= PV(1+r)^n

= 1663(1+5.5%)^19

= 1663(1.055)^19

=1663 × 2.766

=4599.86

Interest = Future value - Present value

= $4599 - $1663

= 2936

4 0

4 years ago

The Monthly Bank pays 3 percent interest, compounded monthly, on their savings accounts. The Daily Bank pays 3 percent interest,

Novosadov [1.4K]

Answer:

The amount that we should deposit in each bank is around $942.

Step-by-step explanation:

Case 1:

A=$1000

n = 12

t = 2

r = 3% or 0.03

p = ?

The compound interest formula is :

$A=p(1+\frac{r}{n})^{nt}$

Substituting values in the formula;

$1000=p(1+\frac{0.03}{12})^{12\times2}$

=> $1000=p(1.0025)^{24}$

=> $1000=1.06175p$

$p=\frac{1000}{1.06175}$

p = $941.84

Case 2:

A=$1000

n = 365

t = 2

r = 3% or 0.03

p = ?

$1000=p(1+\frac{0.03}{365})^{365\times2}$

=> $1000=p(1.0000822)^{730}$

=> $1000=1.06184p$

$p=\frac{1000}{1.06184}$

p = $941.76

The amount that we should deposit in each bank is around $942.

5 0

3 years ago

Read 2 more answers

A contractor is required by a county planning department to submit one, two, three, four, or five forms (depending on the nature

Westkost [7]

Answer:

(a) The value of k is $\frac{1}{15}$ .

(b) The probability that at most three forms are required is 0.40.

(c) The probability that between two and four forms (inclusive) are required is 0.60.

(d) $P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5$ is not the pmf of y.

Step-by-step explanation:

The random variable Y is defined as the number of forms required of the next applicant.

The probability mass function is defined as:

$P(y) = \left \{ {{ky};\ for \ y=1,2,...5 \atop {0};\ otherwise} \right$

(a)

The sum of all probabilities of an event is 1.

Use this law to compute the value of k.

$\sum P(y) = 1\\k+2k+3k+4k+5k=1\\15k=1\\k=\frac{1}{15}$

Thus, the value of k is $\frac{1}{15}$ .

(b)

Compute the value of P (Y ≤ 3) as follows:

$P(Y\leq 3)=P(Y=1)+P(Y=2)+P(Y=3)\\=\frac{1}{15}+\frac{2}{15}+ \frac{3}{15}\\=\frac{1+2+3}{15}\\ =\frac{6}{15} \\=0.40$

Thus, the probability that at most three forms are required is 0.40.

(c)

Compute the value of P (2 ≤ Y ≤ 4) as follows:

$P(2\leq Y\leq 4)=P(Y=2)+P(Y=3)+P(Y=4)\\=\frac{2}{15}+\frac{3}{15}+\frac{4}{15}\\ =\frac{2+3+4}{15}\\ =\frac{9}{15} \\=0.60$

Thus, the probability that between two and four forms (inclusive) are required is 0.60.

(d)

Now, for $P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5$ to be the pmf of Y it has to satisfy the conditions:

$P(y)=\frac{y^{2}}{50}>0;\ for\ all\ values\ of\ y \\$
$\sum P(y)=1$

Check condition 1:

$y=1:\ P(y)=\frac{y^{2}}{50}=\frac{1}{50}=0.02>0\\y=2:\ P(y)=\frac{y^{2}}{50}=\frac{4}{50}=0.08>0 \\y=3:\ P(y)=\frac{y^{2}}{50}=\frac{9}{50}=0.18>0\\y=4:\ P(y)=\frac{y^{2}}{50}=\frac{16}{50}=0.32>0 \\y=5:\ P(y)=\frac{y^{2}}{50}=\frac{25}{50}=0.50>0$

Condition 1 is fulfilled.

Check condition 2:

$\sum P(y)=0.02+0.08+0.18+0.32+0.50=1.1>1$

Condition 2 is not satisfied.

Thus, $P(y)=\frac{y^{2}}{50} ;\ y=1, 2, ...5$ is not the pmf of y.

7 0

3 years ago

Please I need the answer :-)

Arturiano [62]

Answer:

Step-by-step explanation:

5 0

2 years ago

The difference of the square of a number and 12 is equal to 4 times that number. find the negative solution.

dybincka [34]

Im sorry but Im not sure

4 0

3 years ago