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Studentka2010 [4]
3 years ago
5

25 + 14 = blank + 25

Mathematics
1 answer:
sleet_krkn [62]3 years ago
6 0

Answer:

14

Step-by-step explanation:

25 + 14 = 14 + 25

39 = 39

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you deposit $2,000 in an account that earns simple interest. after 6 months the account earns $210 in interest. what is the annu
Luden [163]

Answer:

21%

Step-by-step explanation:

210 is 10.5% of the amount that is for 6 months doubling it will make it 420 that is 21% of the total amount of 2000.

8 0
3 years ago
Given h(×) =-3×+2,solve for x when h (×)=-4
MakcuM [25]

Answer:14

Step-by-step explanation:

h(x)=-3x+2

h(-4)=-3(-4)+2

h(-4)=12+2

h(-4)=14

7 0
4 years ago
Read 2 more answers
Help please ?! I need answers really fast
Aleonysh [2.5K]
The value of x in terms of p is 15/p.

The value of x when p is -5 is -3.

Pls tell me if i’m wrong in any way!
6 0
3 years ago
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
List three ways to express 3^5 as a product of powers
adelina 88 [10]
3^5
3 x 3 x 3
3:5 <- I think
3 0
3 years ago
Read 2 more answers
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