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4vir4ik [10]
3 years ago
15

You need 70 bookmarks for a craft fair you begin with 35 bookmarks and make 10 bookmarks each hour. Write an expression that rep

resents the number of bookmarks you have after h hours
Mathematics
1 answer:
professor190 [17]3 years ago
7 0

Answer:

70 = 10x + 35 is your equation (10x + 35 is your expression).

3 1/2 hr is the time needed.

Step-by-step explanation:

Let hours = x.

You need 70 bookmarks, you have 35 (constant number), and you make 10 bookmarks an hour (varys).

Set the equation:

70 = 10x + 35

Solve for x. Note the equal sign, what you do to one side, you do to the other. Do the opposite of PEMDAS. First subtract, then divide.

Subtract 35 from both sides:

70 (-35) = 10x + 35 (-35)

70 - 35 = 10x

35 = 10x

Isolate the variable (x). Divide 10 from both sides:

(35)/10 = (10x)/10

x = 35/10

x = 3.5

3 1/2 hours is your answer.

~

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A random sample of 36 students at a community college showed an average age of 25 years. Assume the ages of all students at the
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Answer:

98% confidence interval for the average age of all students is [24.302 , 25.698]

Step-by-step explanation:

We are given that a random sample of 36 students at a community college showed an average age of 25 years.

Also, assuming that the ages of all students at the college are normally distributed with a standard deviation of 1.8 years.

So, the pivotal quantity for 98% confidence interval for the average age is given by;

             P.Q. = \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \bar X = sample average age = 25 years

            \sigma = population standard deviation = 1.8 years

            n = sample of students = 36

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So, 98% confidence interval for the average age, \mu is ;

P(-2.3263 < N(0,1) < 2.3263) = 0.98

P(-2.3263 < \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } } < 2.3263) = 0.98

P( -2.3263 \times {\frac{\sigma}{\sqrt{n} } < {\bar X - \mu} < 2.3263 \times {\frac{\sigma}{\sqrt{n} } ) = 0.98

P( \bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} } < \mu < \bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} } ) = 0.98

98% confidence interval for \mu = [ \bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} } , \bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} } ]

                                                  = [ 25 - 2.3263 \times {\frac{1.8}{\sqrt{36} } , 25 + 2.3263 \times {\frac{1.8}{\sqrt{36} } ]

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Therefore, 98% confidence interval for the average age of all students at this college is [24.302 , 25.698].

8 0
3 years ago
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lutik1710 [3]
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3 years ago
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frosja888 [35]

Answer:

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Step-by-step explanation:

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In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

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