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3 years ago

HeeeeeEEEEEELP

1 answer:

5 0

3(x+2)+x=2x+16
3x+6+x=2x+16
4x+6=2x+16
4x=2x+10
2x=10
x=5
The answer is d, x=5
I worked it step by step so I hope you were able to follow

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Multiply (−4)(−2)(−5) A. -40 B. -8 C. 8 D. 40

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(-4)(-2)= 8
8(-5)= -40
Answer: -40

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2 years ago

How do you pass algebra class

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2 years ago

The area of a rectangle is 144 square centimeters. The width is 9 centimeters. Which of the following statements is true? Select

Ket [755]

Answer:

Option C and D are correct.

Step-by-step explanation:

Area of rectangle = 144 cm^2

Width of rectangle = 9 cm

Length of rectangle = ?

We know,

Area of rectangle = Length * Width

144 = Length * 9

144/9 = Length

=> length = 16 cm

Option A is incorrect as 3 times width = 3* 9 = 27 but our length = 16 cm

Option B is incorrect as length = 16 cm and not 63 cm

Option C is correct as Length < 2(Width)

=> 16 < 2(9) => 16 < 18 which is true.

Option D is correct.

Perimeter = 2(Length + Width)

Perimeter = 2(16+9)

Perimeter = 50 cm

Option E is incorrect as Length ≠ Width

6 0

2 years ago

Read 2 more answers

The arrivals of clients at a service firm in Santa Clara is a random variable from Poisson distribution with rate 2 arrivals per

ICE Princess25 [194]

Answer:

1.76% probability that in one hour more than 5 clients arrive

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

The arrivals of clients at a service firm in Santa Clara is a random variable from Poisson distribution with rate 2 arrivals per hour.

This means that $\mu = 2$

What is the probability that in one hour more than 5 clients arrive

Either 5 or less clients arrive, or more than 5 do. The sum of the probabilities of these events is decimal 1. So

$P(X \leq 5) + P(X > 5) = 1$

We want P(X > 5). So

$P(X > 5) = 1 - P(X \leq 5)$

In which

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-2}*2^{0}}{(0)!} = 0.1353$

$P(X = 1) = \frac{e^{-2}*2^{1}}{(1)!} = 0.2707$

$P(X = 2) = \frac{e^{-2}*2^{2}}{(2)!} = 0.2707$

$P(X = 3) = \frac{e^{-2}*2^{3}}{(3)!} = 0.1804$

$P(X = 4) = \frac{e^{-2}*2^{4}}{(4)!} = 0.0902$

$P(X = 5) = \frac{e^{-2}*2^{5}}{(5)!} = 0.0361$

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.1353 + 0.2702 + 0.2702 + 0.1804 + 0.0902 + 0.0361 = 0.9824$

$P(X > 5) = 1 - P(X \leq 5) = 1 - 0.9824 = 0.0176$

1.76% probability that in one hour more than 5 clients arrive

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2 years ago

Read 2 more answers

.A bag contains 3 red checkers and 5 black checkers. A checker is selected, kept out of the bag, and then another checker is sel

cupoosta [38]

Answer: 15 / 56

Step-by-step explanation:

Number of red checkers = 3

Number of black checkers = 5

Total number of checkers = 8

P(black, then red) = 5/8 × 3/7 = 15/56

We should note that the probability to pick a black checker first will be 5 out of 8. Then, there'll be 7 checkers left and the probability to pick a res checker will be 3/7. We then multiply the probabality of each together.

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2 years ago