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3 years ago

13

I NEED HELP what is the answer

1 answer:

guajiro [1.7K]3 years ago

3 0

B, angle 2 is similar/equal to angle 8

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Renting video games from Store S costs $2.50 per game plus a monthly fee of $5.00. Renting video games from Store T costs $5.00

elixir [45]

I believe it is B. Sorry I’m not the best at math but hope this helps !!

6 0

4 years ago

PLZ HELP WORTH 20 POINTS

sergiy2304 [10]

Choice c 24 glad to help explanation: 5 raised to second power is 25 and 25-3 to the zero power is one bc anything to the zero power is one so the answer is 24 25-1 is 24

8 0

4 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

34,8000<br>34.01.<br>3. What is 7,639,586.21 rounded to the nearest ten thousand?

Ket [755]

Answer:

7,640,000

Step-by-step explanation:

The digit 3 is the ten thousands place. However, because the thousands place is 9, you round up to 40,000 to get 7,640,000.

8 0

4 years ago

Pls help me I’ll brainlest

baherus [9]

Answer:

17 3 and 4

Step-by-step explanation:

8 0

3 years ago

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