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Semmy [17]
3 years ago
7

The cost charged for each burger, c, at Bobby's Burger Palace is based on the cost of a plain burger plus an additional charge f

or each topping t. C=0.95t+3.95
Mathematics
1 answer:
Radda [10]3 years ago
7 0

Answer:

C'(t) = \frac{t-3.95}{0.95}

Step-by-step explanation:

Given

C=0.95t+3.95

Required

The inverse function

C=0.95t+3.95

Swap C and t

t=0.95C+3.95

Make C the subject

0.95C = t - 3.95

Divide by 0.95

C = \frac{t}{0.95} - \frac{3.95}{0.95}

Hence, the inverse function is:

C'(t) = \frac{t}{0.95} - \frac{3.95}{0.95}

or

C'(t) = \frac{t-3.95}{0.95}

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3 years ago
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Step-by-step explanation:

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7 0
3 years ago
In a survey of consumers aged 12 and​ older, respondents were asked how many cell phones were in use by the household.​ (No two
bogdanovich [222]

Answer:

The probability that​ his/her household has four or more cell phones in use is 0.122 or 12.2%.

Step-by-step explanation:

<u>The complete question is</u>: In a survey of consumers aged 12 and​ older, respondents were asked how many cell phones were in use by the household.​ (No two respondents were from the same​ household.) Among the​ respondents, 215 answered​ "none", 280 said​ "one", 362 said​ "two", 149 said​ "three," and 140 responded with four or more. A survey respondent is selected at random. Find the probability that​ his/her household has four or more cell phones in use. Is it unlikely for a household to have four or more cell phones in​ use? Consider an event to be unlikely if its probability is less than or equal to 0.05.

We are given that among the​ respondents, 215 answered​ "none", 280 said​ "one", 362 said​ "two", 149 said​ "three," and 140 responded with four or more.

A survey respondent is selected at random.

As we know that the probability of any event is calculated as;

                Probability = \frac{\text{Favorable number of outcomes}}{\text{Total number of outcomes}}

Here, we have to find the probability that​ his/her household has four or more cell phones in use;

Number of respondent having four or more cell phones in use = 140

Total number of respondents asked = 215 + 280 + 362 + 149 + 140 = 1146

<u>So, the required probability</u> = \frac{140}{1146}

                                              = 0.122 or 12.2%

No, it is not unlikely for a household to have four or more cell phones in​ use because the probability is not less than or equal to 0.05 but in actual it is greater than 0.05.

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3 years ago
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Dovator [93]
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8 0
3 years ago
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