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kicyunya [14]
3 years ago
5

Someone please help me with thiss

Mathematics
1 answer:
Olenka [21]3 years ago
7 0

Answer:

The total trip will be 4 hours long.

Step-by-step explanation:

The length of the trip decreases by 65/2 miles for every half hour. So in 2 half hours (or 1 hour), Aubree would have traveled the last 65 miles she needed.

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(5ab - 10b^2 + 15bc) ÷ 5b
Brilliant_brown [7]

Answer:

Final answer: a-2b+3c

Step-by-step explanation:

5ab/5b= a

-10b^2/5b= -2b

15bc/5b= 3c  

7 0
3 years ago
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timurjin [86]

Answer: 3x + (-4) = x + 2

Step-by-step explanation:

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8 0
3 years ago
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Anyone know this pls help
svet-max [94.6K]
9x+3y=12\\
3x+y=4\\
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7 0
3 years ago
A random sample of 10 parking meters in a resort community showed the following incomes for a day. Assume the incomes are normal
GenaCL600 [577]

Answer:

A 95% confidence interval for the true mean is [$3.39, $6.01].

Step-by-step explanation:

We are given that a random sample of 10 parking meters in a resort community showed the following incomes for a day;

Incomes (X): $3.60, $4.50, $2.80, $6.30, $2.60, $5.20, $6.75, $4.25, $8.00, $3.00.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

                         P.Q.  =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean income = \frac{\sum X}{n} = $4.70

            s = sample standard deviation = \sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }  = $1.83

            n = sample of parking meters = 10

            \mu = population mean

<em>Here for constructing a 95% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.</em>

<u>So, 95% confidence interval for the population mean, </u>\mu<u> is ;</u>

P(-2.262 < t_9 < 2.262) = 0.95  {As the critical value of t at 9 degrees of

                                            freedom are -2.262 & 2.262 with P = 2.5%}  

P(-2.262 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 2.262) = 0.95

P( -2.262 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu < 2.262 \times {\frac{s}{\sqrt{n} } } ) = 0.95

P( \bar X-2.262 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+2.262 \times {\frac{s}{\sqrt{n} } } ) = 0.95

<u>95% confidence interval for</u> \mu = [ \bar X-2.262 \times {\frac{s}{\sqrt{n} } } , \bar X+2.262 \times {\frac{s}{\sqrt{n} } } ]

                                         = [ 4.70-2.262 \times {\frac{1.83}{\sqrt{10} } } , 4.70+ 2.262 \times {\frac{1.83}{\sqrt{10} } } ]

                                         = [$3.39, $6.01]

Therefore, a 95% confidence interval for the true mean is [$3.39, $6.01].

The interpretation of the above result is that we are 95% confident that the true mean will lie between incomes of $3.39 and $6.01.

Also, the margin of error  =  2.262 \times {\frac{s}{\sqrt{n} } }

                                          =  2.262 \times {\frac{1.83}{\sqrt{10} } }  = <u>1.31</u>

4 0
3 years ago
Ms.beatly Is trying to get healthy and starts running in the morning.She is able to run 5 miles in 45 minutes how many how many
ElenaW [278]

Answer:

1. 9 minutes

Step-by-step explanation:

1. If she runs 5 miles in 45 minutes, you have 5 miles = 45 minutes. If you divide both sides of this ratio by 5, you get 1 miles = 9 minutes

2. Find out how many each person sells per week:

Emily: 4 boxes = 2 weeks

divide both sides by 2

2 boxes = 1 week

Jose: 9 boxes = 3 weeks

divide both sides by 3

3 boxes = 1 week

This means that jose sells more, because 3 boxes > 2 boxes

6 0
3 years ago
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