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saul85 [17]
3 years ago
6

The cost of labor for servicing cars at B&B Automotive is $50 for each whole hour or for any fraction of an hour. There is a

$25 additional charge to get the oil changed & fluid checked. What would be the cost of servicing a car that required 3.45 hours of labor if the owner requested that the oil be changed and the fluids be checked? a. $100 c. $225 b. $175 d. $250
Mathematics
1 answer:
Masja [62]3 years ago
4 0

Answer:

<h2>d. $250</h2>

Step-by-step explanation:

We can use the equation of a straight line to model the cost of servicing the car.

let the cost be y and the number of hours be x

and the charge per hour is m

y=mx+c

y=50x+25

given that the time is 3.45 hours it is assumed that the charge is for 4 hours since for a fraction of 0.45 hours we are charged $50

y=50(4)+25

y=200+25

y=$225

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Find the values<br>using<br>expansion formula<br>1) 1.5×2.5​
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3.75

Step-by-step explanation:

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Answer:

Step-by-step explanation:

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3 years ago
Read 2 more answers
50 1/2 divided by 1/4 = 202<br> How are the numbers related?
Stels [109]

Answer:

see the explanation

Step-by-step explanation:

Let

x=50\frac{1}{2}=\frac{50*2+1}{2}=\frac{101}{2}

y=\frac{1}{4}

Find the ratio x/y

\frac{x}{y}=\frac{(101/2)}{(1/4)} =\frac{101*4}{2*1}=202

therefore

\frac{x}{y}=202

x=202y

That means----> The first number (50 1/2) is 202 times greater than the second number (1/4)

3 0
3 years ago
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According to a study done by Wakefield Research, the proportion of Americans who can order a meal in a foreign language is 0.47.
UNO [17]

Answer:

Probability that the proportion of Americans who can order a meal in a foreign language is greater than 0.5 is 0.19766.

Step-by-step explanation:

We are given that according to a study done by Wake field Research, the proportion of Americans who can order a meal in a foreign language is 0.47.

Suppose a random sample of 200 Americans is asked to disclose whether they can order a meal in a foreign language.

<em>Let </em>\hat p<em> = sample proportion of Americans who can order a meal in a foreign language</em>

The z-score probability distribution for sample proportion is given by;

          Z = \frac{ \hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion

p = population proportion of Americans who can order a meal in a foreign language = 0.47

n = sample of Americans = 200

Probability that the proportion of Americans who can order a meal in a foreign language is greater than 0.5 is given by = P( \hat p > 0.50)

  P( \hat p > 0.06) = P( \frac{ \hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } > \frac{0.5-0.47}{\sqrt{\frac{0.5(1-0.5)}{200} } } ) = P(Z > 0.85) = 1 - P(Z \leq 0.85)

                                                               = 1 - 0.80234 = <u>0.19766</u>

<em>Now, in the z table the P(Z  </em>\leq <em>x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 0.85 in the z table which has an area of 0.80234.</em>

Therefore, probability that the proportion of Americans who can order a meal in a foreign language is greater than 0.5 is 0.19766.

4 0
3 years ago
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