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3 years ago

11

plzzzz someone help. can anyone answer this questions. please reply if you know. this is the question. To decrease an amount by

21%, multiply by

1 answer:

umka2103 [35]3 years ago

6 0

Multiply the original amount by .79 and this will give you the amount with a 21% discount.

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Ed is on a road trip. He has already traveled 221 miles and is driving at a rate of 61 miles per hour. Which equation could be u

Anarel [89]

Answer:

A

Step-by-step explanation:

because 631 is the total miles, the equation will be equal to 631.

Second, because he is driving at 61 miles/hour, 61 will be the rate, so

61x+b=631

sense he already drove 221 miles, that will be where b is, so

61x+221=631

7 0

2 years ago

Solve -6-(-3x-3)=3(-x-4)+4

Mazyrski [523]

      -6 - (-3x - 3) = 3(-x - 4) + 4
   -6 - 1(3x - 3) = 3(-x) - 3(4) + 4
-6 - 1(3x) + 1(3) = -3x - 12 + 4
   -6 - 3x + 3 = -3x - 8
   -6 + 3 - 3x = -3x - 8
   -3 - 3x = -3x - 8
   + 8     + 8
      5 - 3x = -3x
      + 3x + 3x
      5 = 0x
      0    0
      undefined = x

8 0

3 years ago

Pls help me on this problem as soon as possible(if u don’t know pls do not answer

IgorC [24]

8 verticals i’m pretty sure if not uh yea

4 0

2 years ago

Help ASSAP with this question

mote1985 [20]

(4,4)(-6,4)...notice how ur y coordinates on both of ur points is the same...that means u have a horizontal line with a slope of 0....the equation would be
y = 4.....but in point slope form...not 100% sure

y - y1 = m(x - x1)
slope(m) = 0
using (4,4)...x1 = 4 and y1 = 4
now sub
y - 4 = 0(x - 4) <==

y - y1 = m(x - x1)
slope(m) = 0
using (-6,4)...x1 = -6 and y1 = 4
sub
y - 4 = 0(x - (-6) =
y - 4 = 0(x + 6) <==

6 0

4 years ago

A textbook has 500 pages on which typographical errors could occur. Suppose that there are exactly 10 such errors randomly locat

DiKsa [7]

Answer:

The probability of a selection of 50 pages will contain no errors is 0.368

The probability that the selection of the random pages will contain at least two errors is 0.2644

Step-by-step explanation:

From the information given:

Let q represent the no of typographical errors.

Suppose that there are exactly 10 such errors randomly located on a textbook of 500 pages. Let $\mu$ be the random variable that follows a Poisson distribution, then mean $\mu = \dfrac{10}{500}= 0.02$

and the mean that the random selection of 50 pages will contain no error is $\lambda = 50 \times 0.02 =1$

∴

$Pr(q= 0) = \dfrac{e^{-1} (1)^0}{0!}$

Pr(q =0) = 0.368

The probability of a selection of 50 pages will contain no errors is 0.368

The probability that 50 randomly page contains at least 2 errors is computed as follows:

P(X ≥ 2) = 1 - P( X < 2)

P(X ≥ 2) = 1 - [ P(X = 0) + P (X =1 )] since it is less than 2

$P(X \geq 2) = 1 - [ \dfrac{e^{-1} 1^0}{0!} +\dfrac{e^{-1} 1^1}{1!} ]$

$P(X \geq 2) = 1 - [0.3678 +0.3678]$

$P(X \geq 2) = 1 -0.7356$

P(X ≥ 2) = 0.2644

The probability that the selection of the random pages will contain at least two errors is 0.2644

6 0

3 years ago