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AveGali [126]
2 years ago
15

A city Carnival has an entry fee of $15. Each Carnival ride cost $7.50. Write a Linear Equation that can be used to solve for to

tal spend at the Carnival. In the form y = mx + b
Mathematics
1 answer:
maw [93]2 years ago
6 0
The answer would be Y=7.50x+15
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Answer:

This is a weird assignment

Step-by-step explanation:

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Convert 4 2/3% to a fraction in simplest form.
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Linda purchase a prepaid phone card for $30. Long distance calls cost $.17 a minute using this card when do use her card. Linda
balandron [24]

Answer: her call lasted for 44 minutes

Step-by-step explanation:

Linda purchase a prepaid phone card for $30. This means that the credit on her card is $30.

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7 0
3 years ago
Guys please help me with this question
sashaice [31]

Answer:

Step-by-step explanation:

\frac{g}{f} (x)=\frac{g(x)}{f(x)} \\=\frac{3x^2+x}{4x} \\=\frac{x(3x+1)}{4x} \\=\frac{3x+1}{4}

8 0
3 years ago
A data set includes 103 body temperatures of healthy adult humans having a mean of 98.1 F and a standard deviation of 0.56 F. Co
Ira Lisetskai [31]

Answer:

The 99​% confidence interval estimate of the mean body temperature of all healthy humans is (97.955F, 98.245F).

98.6F is above the upper end of the interval, which means that the sample suggests that the mean body temperature could be lower than 98.6F.

Step-by-step explanation:

The first step is finding the confidence interval

The sample size is 103.

The first step to solve this problem is finding how many degrees of freedom there are, that is, the sample size subtracted by 1. So

df = 103-1 = 102

Then, we need to subtract one by the confidence level \alpha and divide by 2. So:

\frac{1-0.99}{2} = \frac{0.01}{2} = 0.005

Now, we need our answers from both steps above to find a value T in the t-distribution table. So, with 102 and 0.005 in the t-distribution table, we have T = 2.63.

Now, we need to find the standard deviation of the sample. That is:

s = \frac{0.56}{\sqrt{103}} = 0.055

Now, we multiply T and s

M = T*s = 2.63*0.055 = 0.145

For the lower end of the interval, we subtract the mean by M. So 98.1 - 0.145 = 97.955F.

For the upper end of the interval, we add the mean to M. So 98.1 + 0.145 = 98.245F.

The 99​% confidence interval estimate of the mean body temperature of all healthy humans is (97.955F, 98.245F).

98.6F is above the upper end of the interval, which means that the sample suggests that the mean body temperature could be lower than 98.6F.

5 0
3 years ago
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