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nataly862011 [7]
3 years ago
10

A 3-mi cab ride costs $7.30. A 8-mi cab ride costs $15.30. Find a linear equation that models a relationship between cost c and

distance d.
Select one:
a. c = 1.91d + 2.50
b. d = 1.60c + 8.00
c. c = 1.60d + 2.50
d. c = 2.43d + 8.00
Mathematics
1 answer:
ASHA 777 [7]3 years ago
7 0

Since, A 3-mi cab ride costs $7.30 and a 8-mi cab ride costs $15.30.

In the information given above, we can consider these as the coordinate points on the axes.

Let (c, d) be the coordinates , where 'c' represents the total cost and 'd' represents the distance traveled.

Cost is on the 'y' axis and distance is on the 'x' axis.

Now, let the first point on the coordinate axes be (3, 7.30) and the other point is given by (8, 15.30).

For the given points say (x_1,y_1) and (x_2,y_2) , equation of line is given by:

(y-y_1)= m(x-x_1) where m(slope) is given by \frac{y_2-y_1}{x_2-x_1}.

The equation of line for the given points is:

(c-7.30) = (\frac{15.30-7.30}{8-3})(d-3)

(c-7.30) = 1.6(d-3)

c-7.30 = 1.6d-4.8

c-1.6d=2.5

So, c = 1.6d+2.5 is the required linear equation that models a relationship between cost c and distance d.

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8 0
3 years ago
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What is a correct method of solving the equation x/3-6=9 for x? please help thx
stira [4]

Answer:

45

Step-by-step explanation:

\frac{x}{3} -6=9

add 6 on both sides to eliminate it(doing this will get us closer to the variable)

x/3=15

multiply both sides by 3

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hope this helps!

5 0
3 years ago
A.
Katen [24]

Answer:

The answer is D.

Step-by-step explanation:

You have to add both functions together :

f(x) = 25 -  {x}^{2}

g(x) = 5 - x

(f + g)(x) = 25 -  {x}^{2}   + 5 - x

(f + g)(x) = 30 -  {x}^{2}  - x

(f + g)(x) =  -  {x}^{2}  - x  + 30

5 0
3 years ago
The lengths of pregnancies are normally distributed with a mean of 266 days and a standard deviation of 15 days. a. Find the pro
MrMuchimi

Answer:

a. 0.313% (0.003134842261), b. 237.79 days (237.788095878)

Step-by-step explanation:

In this case, the length of pregnancies is a normally distributed variable, with a mean of 266 days, and a standard deviation of 15 days.  

A graph showing the distribution, with regions of interest for the answer, is presented below.

<h3>First Part: Find the probability of a pregnancy lasting 307 days or longer.</h3>

To answer the question regarding <em>the probability of a pregnancy lasting 307 days or longer</em>, it is necessary to calculate what the cumulative probability distribution value is at 307 days. By the way, according to the graph below, 307 days are quite far from the population mean (266 days).

Using the function <em>normaldist(266,15).cdf(307)</em>, from free Desmos software on Internet, we find that, at this length (307 days), the sum of all probabilities for all cases at this value is 99.69%  (0.996865157739).

Considering that the total area of the curve is 1, then <em>the probability of pregnancy lasting 307 days or longer</em> is 1 - 0.996865157739 or 0.003134842261 (or 0.00313), approximately 0.313%, a very low probabilty.

This probability is showed as the "light blue" region at the right extreme of the graph.

<h3>Second Part: Find the length that separates premature babies from those who are not premature.</h3>

To find the length that separates premature babies from those who are not premature, it is a question about <em>find the days related with the probability of 3% (or 0.03)</em> to find such premature babies. So, it is a question of finding a percentile (or 100-quantiles): given the cumulative normal distribution curve, what is the value (length of pregnancies) that represents this 3%.

Using the function <em>quantile(normaldist(266,15), 0.03)</em>, from free Desmos software on Internet, we obtained a value of 237.79 days (237.788095878) for the length of pregnacies of premature babies. In other words, those babies whose mothers have a length of pregnancy <em>lower</em> than 237.79 days are considered premature, or this is "the length that separates premature babies from those who are not premature".

The area below 237.79 days is the blue shaded region in the graph below, at the left extreme of it.

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Answer:

there is no question I'm sorry but I will take brainlyist

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