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FrozenT [24]
3 years ago
6

3. A fast food joint sells 5 boxes of chicken nuggets for $33.90. Another fast food joint sells 4 boxe:

Mathematics
1 answer:
mrs_skeptik [129]3 years ago
3 0
33.90/5= 6.78 or 26.88/4= 6.72. The 4 boxes of chicken fingers has the cheaper unit price
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explain why an circle would have an infinite number of lines of reflection that would carry the circle into itself?
Irina-Kira [14]
Circles will never have a side but they only have 1 line that can be counted as a side.
7 0
3 years ago
Read 2 more answers
What is m pqr? <br> Please helps
LiRa [457]

Answer:

134 degrees

Step-by-step explanation:

Line PQS is 180 degrees. That being said, the addition of <PQR and <SQR equal 180 degrees.

Now you need to set up an equation to find out what angle PQR is. Since there is an X, you need to solve for X. The equation looks like this:

(3x+5) + (x+3) = 180

<u>Solve for x</u>

(3x+5) + (x+3) = 180\\4x+8=180\\4x=172\\x=43

Now plug in 43 where you see x. For this question, you only need to focus on (3x+5)

3(43) + 5 = 134

m<PQR = 134°

Do this every, single, time when you see these questions. Remember that line PQS is 180 degrees, and both of those angles are equal to 180 degrees.

If you want to check if this is true, plug in 43 into our equation we made to see if it equals 180 degrees. <em>If it doesn't equal 180, your equation is incorrect.</em>

<em />

3(43)+5+43+3 = 180

134 + 46 = 180

180 = 180 ✅

4 0
2 years ago
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
Pavel [41]

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

            \mu = population average age

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} } ]

                                                  = [24.302 , 25.698]

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

8 0
2 years ago
7/x+2 + 3x-4/x2+5x+6=9
larisa [96]

Answer:

is this in graph or equation? ???

4 0
2 years ago
Im desperate! please help!
djyliett [7]
I would go with A. An estimate.



This may not be the right answer but Good Luck!
6 0
2 years ago
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