Help plssss its urgent

Daniel paid interest of 1,020 in 5years at12%per annum on a loan. Hw much did he borrow

Marat540 [252]

I = Prt
I=1020
r=0.12 (12% converted to decimal by dividing by 100)
t=5
1020=P(0.12)(5)
1020=P(0.6)
P=1700

5 0

2 years ago

Determine wether 2(x+3)= 2x+6

wariber [46]

It is.

2(x+3)=2x +6
Times 2
2x + 6 =2x+6

Hope this helped,
-Tiara

5 0

3 years ago

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

2 years ago

I will mark you the brainiest for the correct answer, please be correct, Have a good day and take care, thanks. ( VIEW THE IMAGE

makkiz [27]

Mistake in Line 2 : Was in the determining the values for a ,b,c

Mistake in Line 4 : Not taking the square roots of both sides.

Step-by-step explanation:

${a}^{2} + {b}^{2} = {c}^{2}$

First Mistake : Line 2

$a = x \\ b = 6 \\ c = 10$

Second Mistake : Line 4

$136 = {x}^{2} \\ not \\ 136 = x$

Correct Solution :

${x}^{2} + {6}^{2} = {10}^{2} \\ {x}^{2} = {10}^{2} - {6}^{2} \\ {x}^{2} = 100 - 36 \\$

${x}^{2} = 64 \\ \sqrt{ {x}^{2} } = \sqrt{64} \\ x = 6$

3 0

3 years ago

A researcher read a 12-page article in 30 minutes. On

neonofarm [45]

Answer: 2.5 minutes

Explanation: Divide the time by the amount of pages to find the amount of time spent of each page. (30/12=2.5)

3 0

3 years ago