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3 years ago

Figure ABCD is translated down by 6 units:

1 answer:

3 0

The sides would still be the same. If the figure ABCD was a square, it would remain the same because all of the points are equally translated down 6 units. Whatever shape the original figure was, it will remain the same, just in a different location.

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5 – 6x = 8x + 17 and simplify <img src="https://tex.z-dn.net/?f=%5Cfrac%7B-6%7D%7B7%7D" id="TexFormula1" title="\frac{-6}{7}" al

NikAS [45]

Answer:

x = -6/7

Step-by-step explanation:

5 – 6x = 8x + 17

-5 - 5 Subtract 5 from both sides

-6x = 8x + 12

-8x - 8x Subtract 8x from both sides

-14x = 12 Divide both sides by -14

x = -12/14 Simplify

x = -6/7

7 0

3 years ago

Suppose that 40 percent of the drivers stopped at State Police checkpoints in Storrs on Spring Weekend show evidence of driving

lesantik [10]

Answer:

a) 0.778

b) 0.9222

c) 0.6826

d) 0.3174

e) 2 drivers

Step-by-step explanation:

Given:

Sample size, n = 5

P = 40% = 0.4

a) Probability that none of the drivers shows evidence of intoxication.

$P(x=0) = ^nC_x P^x (1-P)^n^-^x$

$P(x=0) = ^5C_0 (0.4)^0 (1-0.4)^5^-^0$

$P(x=0) = ^5C_0 (0.4)^0 (0.60)^5$

$P(x=0) = 0.778$

b) Probability that at least one of the drivers shows evidence of intoxication would be:

P(X ≥ 1) = 1 - P(X < 1)

$= 1 - P(X = 0)$

$= 1 - ^5C_0 (0.4)^0 * (0.6)^5$

$= 1 - 0.0778$

$= 0.9222$

c) The probability that at most two of the drivers show evidence of intoxication.

P(x≤2) = P(X = 0) + P(X = 1) + P(X = 2)

$^5C_0 (0.4)^0 (0.6)^5 + ^5C_1 (0.4)^1 (0.6)^4 + ^5C_2 (0.4)^2 (0.6)^3$

$= 0.6826$

d) Probability that more than two of the drivers show evidence of intoxication.

P(x>2) = 1 - P(X ≤ 2)

$= 1 - [^5C_0 (0.4)^0 (0.6)^5 + ^5C_1 (0.4)^1 (0.6)^4 + ^5C_2 * (0.4)^2 (0.6)^3]$

$= 1 - 0.6826$

$= 0.3174$

e) Expected number of intoxicated drivers.

To find this, use:

Sample size multiplied by sample proportion

n * p

= 5 * 0.40

= 2

Expected number of intoxicated drivers would be 2

7 0

3 years ago

X + 3 = x + 6<br> is it one solution no solution or infinitely solutions

mars1129 [50]

Answer:

No solutions I think.

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

In a classroom of 33

amid [387]

Answer: 9 boys
Explanation:
3/11= x/33
11x= 99
x=9

5 0

3 years ago

Three cards are drawn at random from a standard deck of cards, without replacement. Find the probability that the suits of the c

Evgesh-ka [11]

The probability that suits of cards are spades, hearts, and spades is 39/850 or 0.046.

According to the give question.

Three cards are drawn from a standard deck of cards without replacement.

Since,

The total number of cards in a standard deck = 52

Total number of spade cards = 13

Total number of heart cards = 13

Now,

The probabbility of getting first card as a spade card = 13/52 = 1/4

The probability of getting second card as heart card

= 13/(52 -1) (cards are drawn without replacement)

= 13/51

The probability of getting third card as a spade card

= (13 - 1)/52 -2

= 12/50

= 6/25

Therefore, the probability that suits of the cards are spades, hearts, spades

= 1/4 × 13/51 × 6/25

= 39/850

= 0.046

Hence, the probability that suits of cards are spades, hearts, and spades is 39/850 or 0.046.

Find out more information about probability here:

brainly.com/question/11234923

#SPJ4

8 0

1 year ago