Parallelogram MNPQ was dilated to create parallelogram M'N'P'Q'.

Describe in detail how you would plot the value 2 3/8 on a number line

malfutka [58]

Between 2 and 3, after the 2 1/4 point, but before the halfway point.

5 0

4 years ago

Which angles are consecutive interior angles? Check all that apply.

serious [3.7K]

The answer is A because 6+1 is a even number

6 0

3 years ago

From a standard deck of cards, a 5-card hand is dealt. Calculate the number of hands that contain 4 diamonds and 1 black king

rjkz [21]

Answer:

1430

Step-by-step explanation:

Given that a standard deck of cards is 52 cards, we have 13 Diamonds and we need 4 diamonds.

Hence, we have 13C4.

Also, we have 4 kings in a standard deck of cards, and in which we have 2 black kings but we need 1 king.

Hence we have a 2C1

Therefore:

13C4 * 2C1

=> 13! ÷ [4! (13 - 4)!] * 2! ÷ [1! (2-1)!]

=> 13! ÷ [4! (9)!]

=> (1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10 x 11 x 12 x 13) ÷ [(1 x 2 x 3 x 4) (1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9)]

=> (10 x 11 x 12 x 13) ÷ 24

=> (17160 ÷ 24) * (24 ÷ 6)

=> 715

2! ÷ [1! (2-1)!]

=> 2

Hence, we have 715 * 2

=> 1430

Hence, in this case, the correct answer to the question is 1430

7 0

3 years ago

Automobiles arrive at a vehicle equipment inspection station according to a Poisson process with rate α = 8 per hour. Suppose th

Elis [28]

Answer:

a) The probability that exactly eight arrive during the hour and all eight have no violations is 0.0005.

b) For any fixed y ≥ 8, the probability that y arrive during the hour, of which eight have no violations is:

$P(X=y\,\&\,nv=8)=\frac{4^ye^{-8}}{8!(y-8)!}$

c) The probability that eight "no-violation" cars arrive during the next hour is 0.030.

Step-by-step explanation:

a) The probability that exactly eight arrive during the hour and all eight have no violations is equal to the product of the probability of arrival of 8 vehicules and the probability of having 8 vehicules with no violations.

$P(X=8\,\&\,no\, violations)=P(no\, violations|X=8)*P(X=8)\\\\P(X=8\,\&\,nv)=(0.5)^8*\frac{8^8e^{-8}}{8!} = 0.0039*0.1396=0.0005$

b) For any fixed y ≥ 8, the probability that y arrive during the hour, of which eight have no violations is:

$P(X=y\,\&\,nv=8)=P(nv=8|X=y)*P(X=y)\\\\P(X=y\,\&\,nv=8)=[\binom{y}{8}(0.5)^8*(0.5)^{y-8}]*\frac{8^ye^{-8}}{y!} =\frac{y!}{8!(y-8)!}0.5^y *8^y*\frac{e^{-8}}{y!}\\\\ P(X=y\,\&\,nv=8)=(\frac{y!}{y!})(0.5*8)^y\frac{e^{-8}}{8!(y-8)!}=\frac{4^ye^{-8}}{8!(y-8)!}$

c) Using the result of point (b) we can express the probability that eight "no violation" vehicules arrive durting the next hour as:

$P(nv=8)=\sum\limits^\infty_{y=8} {\frac{4^ye^{-8}}{8!(y-8)!}}=\frac{e^{-8}}{8!} \sum\limits^\infty_{y=8} {\frac{4^y}{(y-8)!}}=\frac{e^{-8}4^{8}}{8!} \sum\limits^\infty_{y=8} {\frac{4^{y-8}}{(y-8)!}}\\\\P(nv=8)=\frac{e^{-8}4^{8}}{8!} \sum\limits^\infty_{z=0} {\frac{4^{z}}{z!}}=\frac{e^{-8}4^{8}}{8!}*e^4=\frac{e^{-4}4^{8}}{8!}\\\\P(nv=8)= 0.030$

5 0

3 years ago

Hi I’m doing IXL and I’m having trouble with this one

maria [59]

Answer:

They are not equivalent.

Step-by-step explanation:

5:2 = 2,5

8:3 = 2,66...

8 0

3 years ago

Read 2 more answers