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MA_775_DIABLO [31]
3 years ago
13

How to fill out the squares

Mathematics
2 answers:
Lady bird [3.3K]3 years ago
7 0
Compare it to the same graph as the other one
bixtya [17]3 years ago
6 0
Just compare it as a regular graph.:)
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Carla had $100 in her account in May. How much money does she have in her account in August?
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Carla still has $100 because she didn't get any more money in the other months
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3 years ago
Read 2 more answers
Simplify the answer as much as possible​
RSB [31]

Answer:

f(-5)=240

g(-7)=-33

Step-by-step explanation:

1) f(x)=3x^2 - 3x

f(-5)= 3×(-5)^2 - 3×(-5) = -15^2 + 15= 225 + 15=240

2) g(x)=5x + 2

g(-7)= 5×(-7) + 2= -35+2=-33

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3 years ago
[20 POINTS] A cylinder has a height of 11 centimeters and its circle bases have a radius of 9 centimeters. Find the surface area
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A 2 level 5 factor experiment is being conducted to optimize the reliability of an electronic control module. In a full factoria
allsm [11]

Answer:

32

Step-by-step explanation:

In a full factorial experiment, there are usually 32 or 2⁵ treatments. This implies that there are usually  2 levels and 5 factors. From the description below, there is a list of the 32 treatments.

Factor A      Factor B      Factor C      Factor B      Factor E

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3 0
3 years ago
1. In order to get more female customers, a new clothing store offers free gourmet coffee and pastry to its customers. The avera
ioda

Answer:

No, the manager is not correct based on the 95% confidence interval.

Step-by-step explanation:

We are given that the average daily revenue over the past five-week period has been $1,080 with a standard deviation of $260, i.e.; X bar = $1080 and s = $260 and sample size, n = 35 .

The Pivotal quantity for 95% confidence interval is given by;

                \frac{Xbar - \mu}{\frac{s}{\sqrt{n} } } ~ t_n_-_1

where, X bar = sample mean = $1080

                s  = sample standard deviation = $260

                 n = sample size = 35 {five-week}

So, 95% confidence interval for average daily revenue, \mu is given by;

P(-2.032 < t_3_4 < 2.032) = 0.95

P(-2.032 < \frac{Xbar - \mu}{\frac{s}{\sqrt{n} } } < 2.032) = 0.95

P(-2.032 * {\frac{s}{\sqrt{n} } < {Xbar - \mu} < 2.032 * {\frac{s}{\sqrt{n} } ) = 0.95

P(X bar - 2.032 * {\frac{s}{\sqrt{n} } < \mu < X bar + 2.032 * {\frac{s}{\sqrt{n} } ) = 0.95

95% confidence interval for \mu = [ X bar - 2.032 * {\frac{s}{\sqrt{n} } , X bar + 2.032 * {\frac{s}{\sqrt{n} } ]

                                            = [ 1080 - 2.032 * {\frac{260}{\sqrt{35} } , 1080 + 2.032 * {\frac{260}{\sqrt{35} } ]

                                             = [ 990.70 , 1169.30 ]

<em>No, the manager is not correct based on the fact that the coffee and pastry strategy would lead to an average daily revenue of $1,200 because the calculate 95% confidence interval does not include value of $1200.</em>

Therefore, the store manager believe is not correct.

8 0
3 years ago
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