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4 years ago

Examine the following graph on the line

Mathematics

1 answer:

Lorico [155]4 years ago

6 0

Answer:

B. y - 0 = -⅓(x + 2)

Step-by-step explanation:

First, find the slope of the line using two points in the line, (-5, 1) and (-2, 0):

Slope (m) = ∆y/∆x = -⅓

To find the equation in point-slope form, substitute (a, b) = (-2 0), and m = -⅓ into y - b = m(x - a)

Thus:

y - 0 = -⅓(x -(-2))

y - 0 = -⅓(x + 2)

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3 years ago

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A survey conducted by the Consumer Reports National Research Center reported, among other things, that women spend an average of

Nookie1986 [14]

Answer:

(a) The probability that a randomly selected woman shop exactly two hours online is 0.217.

(b) The probability that a randomly selected woman shop 4 or more hours online is 0.0338.

Step-by-step explanation:

Let X = time spent per week shopping online.

It is provided that the random variable X follows a Poisson distribution.

The probability function of a Poisson distribution is:

$P (X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!} ;\ x=0,1,2,...$

The average time spent per week shopping online is, λ = 1.2.

(a)

Compute the probability that a randomly selected woman shop exactly two hours online over a one-week period as follows:

$P (X=2)=\frac{e^{1.2}(1.2)^{2}}{2!} =0.21686\approx0.217$

Thus, the probability that a randomly selected woman shop exactly two hours online is 0.217.

(b)

Compute the probability that a randomly selected woman shop 4 or more hours online over a one-week period as follows:

P (X ≥ 4) = 1 - P (X < 4)

= 1 - P (X = 0) - P (X = 1) - P (X = 2) - P (X = 3)

$=1-\frac{e^{1.2}(1.2)^{0}}{0!}-\frac{e^{1.2}(1.2)^{1}}{1!}-\frac{e^{1.2}(1.2)^{2}}{2!}-\frac{e^{1.2}(1.2)^{2}}{3!}\\=1-0.3012-0.3614-0.2169-0.0867\\=0.0338$

Thus, the probability that a randomly selected woman shop 4 or more hours online is 0.0338.

(c)

Compute the probability that a randomly selected woman shop less than 5 hours online over a one-week period as follows:

P (X < 5) = P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)

$=\frac{e^{1.2}(1.2)^{0}}{0!}+\frac{e^{1.2}(1.2)^{1}}{1!}+\frac{e^{1.2}(1.2)^{2}}{2!}+\frac{e^{1.2}(1.2)^{3}}{3!}+\frac{e^{1.2}(1.2)^{4}}{4!}\\=0.3012+0.3614+0.2169+0.0867+0.0260\\=0.9922$

Thus, the probability that a randomly selected woman shop less than 5 hours online is 0.9922.

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4 years ago

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3 years ago

______% of 200 miles is 150 miles.

Genrish500 [490]

Answer:

75%

Step-by-step explanation:

150/200 × 100 =75

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3 years ago

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a recycling bin is in the shape of a rectangular box. find the height of the box if its length is 16ft, its width is 6 ft, and i

Andre45 [30]

Answer:

Height 4 ft

Step-by-step explanation:

Surface area= 2(lw + lh + wh) =

2(16)16)+2(16h)+2(6h) =

2(16)(6)+32h+12h=368

192 + 44h = 368

44h = 368 - 192

44h = 176

h = 176/44

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3 years ago