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3 years ago

I need help with question 6

Mathematics

2 answers:

Vilka [71]3 years ago

7 0

A. -14
B. 0
C. 6
D. 0

so the answer is D. 6
hope this helped

uranmaximum [27]3 years ago

4 0

B!!! Because they A equals -20, B equals 20, C equals 0, and D equals 0!

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The mean of five numbers is 6 the median is 7 and the mode is 2. What is the answer?

Lady_Fox [76]

Answer:

2,2,7,9,10

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3 years ago

Is lmn congruent to opq if so name the postulate

insens350 [35]

Answer:

Option (A)

Step-by-step explanation:

Given:

LM ≅ OP

MN ≅ PQ

∠M ≅ ∠P

To Prove:

ΔLMN ≅ ΔOQP

Statements Reasons

1). LM ≅ OP 1). Given

2). MN ≅ PQ 2). Given

3). ∠P ≅ ∠M 3). Given

4). ΔLNM ≅ ΔOQP 4). By the SAS postulate of congruence.

[Side - Angle - Side]

Therefore, Option (A) will be the answer.

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4 years ago

Is 0.6 60 and 6 equal?

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Answer:

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Step-by-step explanation:

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2 years ago

Read 2 more answers

Solve X/3 - 2X+1/3 = X-3/5

iren2701 [21]

Answer:

Answer is 3

Step-by-step explanation:

x/3-2x/1+3 =x-3/5

x/3-x/3 = x-3/5

0=x-3/5

0=x-3

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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