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Gnoma [55]
3 years ago
8

Lella will rent a car for the weekend. She can choose one of two plans. The first plan has an initial fee of $46 and costs an ad

ditional $0.14 per mile driven. The
second plan has an initial fee of $51 and costs an additional $0.10 per mile driven

Mathematics
1 answer:
Helen [10]3 years ago
6 0
The plans cost the same at 125 miles. the cost is $63.50
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Solve the system by linear combination. Show all your work.<br><br> 9x+5y=35<br> -2x-5y=0
mars1129 [50]

Answer:

(5, - 2 )

Step-by-step explanation:

Given the 2 equations

9x + 5y = 35 → (1)

- 2x - 5y = 0 → (2)

Adding the 2 equations term by term will eliminate the y- term

7x = 35 ( divide both sides by 7 )

x = 5

Substitute x = 5 into either of the 2 equations for value of y

Substituting in (1) gives

9(5) + 5y = 35

45 + 5y = 35 ( subtract 45 from both sides )

5y = - 10 ( divide both sides by 5 )

y = - 2

Solution is (5, - 2 )

5 0
3 years ago
Help it's so hard for me​
jenyasd209 [6]

Answer:

the answer is 11.9 pounds

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
A machine is used to fill 1-liter bottles of a type of soft drink. We can assume that the output of the machine approximately fo
allsm [11]

Answer:

1.00434

Step-by-step explanation:

Given the following :

Given a normal distribution ;

Mean (m) = 1.0 liter

Standard deviation (σ) = 0.01 liter

Sample size (n) = 25

For 97% sample means (sm) = 0.97

Z = (m - sm) / s

Zcrit = 1 - (100% - 97%)/2

Zcrit = 1 - (0.03/2)

Zcrit = 1 - 0.015 = 0.985

The z score which corresponds to 0.985 = 2.17

Upper limit : m + z*(σ/√n)

Upper limit : 1.0 + 2.17*(0.01/√25)

Upper limit : 1. 0 + 2.17*(0.01/5)

= 1.0 + 2.17*0.002

= 1.0 + 0.00434

= 1.00434

3 0
3 years ago
What percentage of the birds shown in the pie chart are notgreen?
Gre4nikov [31]
A 80% would be your answer! 
3 0
3 years ago
Read 2 more answers
A walk-in medical clinic believes that arrivals are uniformly distributed over weekdays (Monday through Friday). It has collecte
allsm [11]

Answer:

The degrees of freedom are given by;

df =n-1= 5-1=4

The significance level is 0.1 so then the critical value would be given by:

F_{cric}= 7.779

If the calculated value is higher than this value we can reject the null hypothesis that the arrivals are uniformly distributed over weekdays

Step-by-step explanation:

For this case we have the following observed values:

Mon 25 Tue 22 Wed 19 Thu 18 Fri 16 Total 100

For this case the expected values for each day are assumed:

E_i = \frac{100}{5}= 20

The statsitic would be given by:

\chi^2 = \sum_{i=1}^n \frac{(O_i-E_i)^2}{E_i}

Where O represent the observed values and E the expected values

The degrees of freedom are given by;

df =n-1= 5-1=4

The significance level is 0.1 so then the critical value would be given by:

F_{cric}= 7.779

If the calculated value is higher than this value we can reject the null hypothesis that the arrivals are uniformly distributed over weekdays

5 0
3 years ago
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