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stellarik [79]
3 years ago
11

Rewrite 2+(1/b-2) = (3b/b+2) as a proportion. Which of these proportions is equivalent to the original equation?

Mathematics
2 answers:
Makovka662 [10]3 years ago
8 0

Answer:

C for the first one

1,6 for the second

Neither for the third

Step-by-step explanation: EDGE 2021

lorasvet [3.4K]3 years ago
7 0

Answer: c for the first part

1,6 for the second

nether for the third

Step-by-step explanation:

all on edg

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Equilateral triangle X Y W has degree angles of 60 degrees. Point D extends from side X W. Angle D X Y is 120 degrees. Triangle
V125BC [204]

Answer:

80

Step-by-step explanation:

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3 years ago
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The service department of a luxury car dealership conducted research on the amount of time its service technicians spend on each
mart [117]

Answer:

Probability that the mean service time is between 1 and 2 hours is 0.96764.

Step-by-step explanation:

We are given that a systematic random sample of 100 service appointments has been collected.

The 100 appointments showed an average preparation time of 90 minutes with a standard deviation of 140 minutes.

<u><em>Let </em></u>\bar X<u><em> = sample mean service time</em></u>

The z-score probability distribution for sample mean is given by;

                             Z = \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \mu = average preparation time = 90 minutes

           \sigma = standard deviation = 140 minutes

           n = sample of appointments = 100

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, probability that the mean service time is between 60 and 120 minutes is given by = P(60 minutes < \bar X < 120 minutes)

P(60 minutes < \bar X < 120 minutes) = P(\bar X < 120 min) - P(\bar X \leq 60 min)  

  P(\bar X < 120 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{120-90}{\frac{140}{\sqrt{100} } } ) = P(Z < 2.14) = 0.98382

  P(\bar X \leq 60 min) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } \leq \frac{60-90}{\frac{140}{\sqrt{100} } } ) = P(Z \leq -2.14) = 1 - P(Z < 2.14)

                                                        = 1 - 0.98382 = 0.01618

<em>The above probability is calculated by looking at the value of x = 2.14 in the z table which has an area of 0.98382.</em>

Therefore, P(60 min < \bar X < 120 min) = 0.98382 - 0.01618 = <u>0.96764</u>

7 0
3 years ago
Find the area of the composite figure. (round to the nearest hundredth)
8090 [49]
We know that
[the area of composite figure]=[area of rectangle]+[area of a half circle]

[area of rectangle]=b*h
where
b=the base
h=the height 
[area of rectangle]=b*h--------> 4.1*8.6---------> 35.26 cm²

[area of a circle]=pi*r²
where 
r=the radius
r=4.1/2-------> 2.05 cm
the figure is a half circle
then 
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[the area of composite figure]=35.26+6.60--------> 41.86 cm²

the answer is
41.86 cm² 

8 0
3 years ago
If h(x) = 5 + x and k (x) = StartFraction 1 Over x EndFraction, which expression is equivalent to (k circle h) (x)?
Semenov [28]

Answer:

The answer is B 1/(5+x)

Step-by-step explanation: I just took this test on edgenuity and had trouble with this exact question

8 0
3 years ago
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