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Lostsunrise [7]
3 years ago
6

A factory produces 40 cars per week. What is the percentage in the production rate if it changes from 40 cars per week to 100 ca

rs per week?
Mathematics
2 answers:
zzz [600]3 years ago
5 0
250% because it multiplies by 2.5
Travka [436]3 years ago
3 0
Answer:

100/40=2.5 or 250%

250% in production rate should be your answer.
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Which could be the length of the third side of a triangle with side lengths 3ft and 9ft? explain
lakkis [162]
The third side can be 12ft
7 0
3 years ago
4x27-3+5x2=......................
kolbaska11 [484]

Answer:

115

Step-by-step explanation:

8 0
3 years ago
What can be concluded about the line represented in the table? Check all that apply. x y –6 –7 2 –3 8 0 The slope is 2. The slop
Stels [109]
The first thing we must do for this case is to find the equation of the line.
 y-yo = m (x-xo)

 We have then:
 m = (y2 - y1) / (x2-x1)

m = (-3 - 0) / (2-8)

m = 1/2
 We choose an ordered pair:
 (xo, yo) = (8, 0)

 Substituting values:
 y-0 = (1/2) (x-8)

y = (1/2) x - 4
 
 From here we conclude:

 Intersection with y:
 
We evaluate x = 0 in the function:
 y = (1/2) (0) - 4

y = -4

 Slope of the line:
 m = 1/2


 Point (-2, -5):
 
We evaluate the value of x = -2 and the value of y = -5
 -5 = (1/2) (- 2) - 4

-5 = -1 - 4

-5 = - 5
 The equation is satisfied.

 Point (8, 0):
 
It is part of the table, therefore belongs to the line.

 Answer:
 
The slope is 1/2
 
The y-intercept is -4.
 
The points (-2, -5) and (8, 0) are also on the line.
7 0
3 years ago
Read 2 more answers
NEED HELP ASAP!!! What is the interquartile range of the data set? Enter the answer in the box. Key: 1|5 means 15 A stem-and-lea
monitta
To find the interquartile range, you will list the data that is presented in the stem and leaf plot.

Find the median of the data (30.5)

Find the median of the lower half and the median of the upper half.

Subtract these two values.

The data are <u>20</u>, 25, 30, 30, 31, 40, 41, <u>49</u>.
                            27.5                40.5
40.5-27.5 = 13

The interquartile range is 13.
3 0
3 years ago
Read 2 more answers
The miles-per-gallon rating of passenger cars is a normally distributed random variable with a mean of 33.8 mpg and a standard d
EleoNora [17]

Answer:

The probability that a randomly selected passenger car gets more than 37.3 mpg is 0.1587.

Step-by-step explanation:

Let the random variable <em>X</em> represent the miles-per-gallon rating of passenger cars.

It is provided that X\sim N(\mu=33.8,\ \sigma^{2}=3.5^{2}).

Compute the probability that a randomly selected passenger car gets more than 37.3 mpg as follows:

P(X>37.3)=P(\frac{X-\mu}{\sigma}>\frac{37.3-33.8}{3.5})

                   =P(Z>1)\\\\=1-P(Z

Thus, the probability that a randomly selected passenger car gets more than 37.3 mpg is 0.1587.

7 0
3 years ago
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