All categories

3 years ago

Explain why I can’t kendo own is jwjnbej Jenner jw whaujn knowwww

Mathematics

1 answer:

makvit [3.9K]3 years ago

5 0

Answer:

‏‏‎ ‏‏‎ ‏‏‎ ‏‏‎ ‏‏‎ ‏‏‎ ‏‏‎ ‏‏‎ ‏‏‎ ‏‏‎ ‏‏‎ ‏‏‎ ‏‏‎ ‏‏‎ ‏‏‎

You might be interested in

a daily newspaper has 10225 subscribers when it began publication. six years old later it has 8200 subscribers what is the avera

seraphim [82]

Answer:

The rate of change in number of subscriber for the six years is 3.62%

Step-by-step explanation:

Given as :

The initial subscriber of newspaper = p = 10225

The subscriber of newspaper after 6 years of publish = P = 8200

The time period = t = 6 years

Let The average yearly rate of change = r%

<u>Now, According to question</u>

The subscriber of newspaper after n years = initial subscriber × $(1-\dfrac{\textrm rate}{100})^{\textrm time}$

Or, P = p × $(1-\dfrac{\textrm r}{100})^{\textrm t}$

Or, 8200 = 10225 × $(1-\dfrac{\textrm r}{100})^{\textrm 6}$

Or, $\dfrac{8200}{10225}$ = $(1-\dfrac{\textrm r}{100})^{\textrm 6}$

Or, 0.80195 = $(1-\dfrac{\textrm r}{100})^{\textrm 6}$

<u>Taking power $\dfrac{1}{6}$ both side</u>

So, $(0.80195 )^{\frac{1}{6}}$ = $((1 - \frac{r}{100} )^{6})^{\frac{1}{6}}$

Or, 0.9638 = 1 - $\dfrac{r}{100}$

Or, $\dfrac{r}{100}$ = 1 - 0.9638

Or, $\dfrac{r}{100}$ = 0.0362

Or, r = 0.0362 × 100

i.e r = 3.62

So, The rate of change in subscriber = 3.62%

Hence, The rate of change in number of subscriber for the six years is 3.62% . Answer

4 0

3 years ago

Two Parallel lines are cut by a transcersal and form a pair of alternate exterior angles. One angle measures (6x + 5) and the ot

zaharov [31]

Answer:

11° is the measure of the exterior angles

Step-by-step explanation:

Theorem: Alternate exterior angles are congruent when two parallel lines cut by a transversal.

First we solve for x using the theorem

6x + 5 = 7x + 4

x = 1

Substitute x = 1 into 6x + 5

6(1) + 5 = 6 + 5 = 11°

4 0

3 years ago

Help ASAP solve: -3/4n = -6

anyanavicka [17]

Answer:

$\huge\boxed{n=8}$

Step-by-step explanation:

$-\dfrac{3}{4}n=-6\qquad|\text{multiply both sides by (-4)}\\\\(-4\!\!\!\!\diagup)\left(-\dfrac{3}{4\!\!\!\!\diagup}n\right)=(-4)(-6)\\\\3n=24\qquad|\text{divide both sides by 3}\\\\\dfrac{3\!\!\!\!\diagup n}{3\!\!\!\!\diagup}=\dfrac{24\!\!\!\!\!\diagup}{3\!\!\!\!\diagup}\\\\n=8$

3 0

3 years ago

Read 2 more answers

What is 0.512 as a percentage

Sonja [21]

Answer:

51.2%

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Ship collisions in the Houston Ship Channel are rare. Suppose the number of collisions are Poisson distributed, with a mean of 1

alexandr1967 [171]

Answer:

$a) \simeq 0.3012$ $b) \simeq 0.0494$ c) $\simeq 0.2438$

Step-by-step explanation:

Rate of collision,

1.2 collisions every 4 months

or, $\frac{1.2}{4}$

= 0.3 collisions per month

So, the Poisson distribution for the random variable no. of collisions per month (X) is given by,

P(X =x) = $\frac{e^{-\lambda}\times {\lambda}^{x}}}{x!}
$

for x ∈ N ∪ {0}

= 0 otherwise --------------------------------------(1)

here, $\lambda = 0.3$ collision / month

No collision over a 4 month period means no collision per month or X =0

Putting X = 0 in (1) we get,

P(X = 0) = $\frac{e^{-0.3}\times {\0.3}^{0}}{0!}
$

$\simeq 0.7408182207$ ------------------------------------(2)

Now, since we are calculating this for 4 months,

so, P(No collision in 4 month period)

= $0.7408182207^{4}$

$\simeq 0.3012$ -----------------------------------------------------------(3)

2 collision in 2 month period means 1 collision per month or X =1

Putting X =1 in (1) we get,

P(X =1) = $\frac{e^{-0.3}\times {\0.3}^{1}}{1!}
$

$\simeq 0.2222454662$ ------------------------------------(4)

Now, since we are calculating this for 2 months, so ,

P(2 collisions in 2 month period)

= $0.2222454662^{2}$

$\simeq 0.0494$ -----------------------------------------(5)

1 collision in 6 months period means

$\frac{1}{6}$ collision per month

Now, P(1 collision in 6 months period)

= P( X = 1/6] (which is to be estimated)

= $\frac {P(X=0)\times 5 + P(X =1)\times 1}{6}$

= \frac {0.7408182207 \times 5 + 0.2222454662 \times 1}{6}[/tex]

$\simeq 0.6543894283$ -------------------------------------------(6)

So,

P(1 collision in 6 month period)

= $0.6543894283^{6}$

$\simeq 0.0785267444$ ------------------------------------------------(7)

So,

P(No collision in 6 months period)

= $(P(X =0)^{6}$

$\simeq 0.1652988882$ ---------------------------------(8)

so,

P(1 or fewer collision in 6 months period)

= (8) + (7 ) = 0.0785267444 +0.1652988882

$\simeq 0.2438$ ---------------------------------------------(9)

7 0

3 years ago