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Pavel [41]
2 years ago
8

You want to buy some noodles. An 8-ounce package costs $2.88. A 14-ounce package costs

Mathematics
1 answer:
torisob [31]2 years ago
6 0

Answer:

A 20-ounce package that costs $6.80 is the best buy

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(5/10)^4 in exponential form
Whitepunk [10]

Answer:

6.25 \times  {10}^{ - 2}

Step-by-step explanation:

{ \bigg( \frac{5}{10} \bigg) }^{4}  \\  \\  =  {(0.5)}^{4}  \\  \\  = 0.0625 \\  \\  = 6.25 \times  {10}^{ - 2}

6 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
2 years ago
A number divided by 35 has a quotient of 8 with a remainder of 9? What is the number?
marissa [1.9K]

Answer: 311.5


Step-by-step explanation: 35*8.9=311.5

311.5/35=8.9


7 0
3 years ago
Read 2 more answers
HELP I WILL MARK BRAINLEST
OverLord2011 [107]

its only showing "help i will mark brainliest"..?

8 0
3 years ago
3. Polygon ABCDE is congruent to polygon TVSRK. EAB corresponds to _____.
mars1129 [50]

Answer: The answer is (A) ∠T.

Step-by-step explanation:  Given that the polygon ABCDE is congruent to the polygon TVSRK. We are to find the corresponding angle of ∠EAB.

In the two polygons, the corresponding vertices are

A ⇒ T

B ⇒ V

C ⇒ S

D ⇒ R

E ⇒ K.

Therefore, ∠EAB, which is ∠A will correspond to ∠T.

Thus, the correct option is (A) ∠T.

6 0
3 years ago
Read 2 more answers
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