All categories

4 years ago

I don't understand can you help me solve it?

Mathematics

1 answer:

Hitman42 [59]4 years ago

4 0

$0.\overline{282}=0.282282282...\\\\x=0.282282282...\ \ \ \ \ \ |multiply\ both\ sides\ by\ 1000\\\\1000x=282.282282282...\\\\1000x-x=282.282282282...-0.282282282...\\\\999x=282\ \ \ \ \ |divide\ both\ sides\ by\ 999\\\\x=\dfrac{282}{999}\\\\x=\dfrac{282:3}{999:3}\\\\x=\dfrac{94}{333}\\\\Answer:\boxed{0.\overline{282}=\frac{94}{333}}$

You might be interested in

How much money do you save when you buy a $40 video game that is on sale for 25% off?

nexus9112 [7]

I it is think it’s $10

4 0

3 years ago

Read 2 more answers

Niech ktoś pomoże ! a)5 2\7 +1 3\7 -1\7= b)4 5\9 -2 4\9 +3 2\9= c)2 2\3 -(1 4\9 +2 3\9=

kvv77 [185]

A) 6 4/7

B) 5 3/9

C) 6 6/3

5 0

4 years ago

HELP PLS PLS PLSP PSLLS

STALIN [3.7K]

Answer:

110

Step-by-step explanation:

5 0

3 years ago

Solve for Y I need help please

muminat

Bring y to one side and the rest to the other switching the positives to negative then you’ll get Y=114

6 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.

(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