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Anna007 [38]
2 years ago
14

The Patel family went apple picking. The number of red apples they picked was 2 __ 2 9 times the number of green apples they pic

ked. If they picked 45 green apples, how many red apples did they pick?
Mathematics
1 answer:
miskamm [114]2 years ago
6 0

Answer:

100

I first made 45 a fraction = 45/1

Then i made 2 2/9 into an improper fraction

then make my equation

45/1 ×20/9=100

answer is 100 red apples

Step-by-step explanation:

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The area of rectangle C is 2

Step-by-step explanation:

Using a scale factor of \frac{1}{5} yields the side lengths of 1 and 2

(1)(2)=2

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Determine which equations below, when combined with the equation 3x-4y=2, will form a system with no solutions. Choose all that
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Two equations will not have solution if they are parallel and have different y-intercepts. Parallel lines have the same slope. In a slope-intercept form, the equation of the line can be expressed as,       
 
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g An irate student complained that the cost of textbooks was too high. He randomly surveyed 36 other students and found that the
blagie [28]

Answer:

A 90% confidence interval of the true mean is [$119.86, $123.34].

Step-by-step explanation:

We are given that an irate student complained that the cost of textbooks was too high. He randomly surveyed 36 other students and found that the mean amount of money spent on textbooks was $121.60.

Also, the standard deviation of the population was $6.36.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

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

where, \bar X = sample mean amount of money spent on textbooks = $121.60

            \sigma = population standard deviation = $6.36

            n = sample of students = 36

            \mu = population mean

<em>Here for constructing a 90% confidence interval we have used One-sample z-test statistics as we know about population standard deviation.</em>

<em />

So, 95% confidence interval for the population mean, \mu is ;

P(-1.645 < N(0,1) < 1.645) = 0.90  {As the critical value of z at 5% level

                                                      of significance are -1.645 & 1.645}  

P(-1.645 < \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } < 1.645) = 0.90

P( -1.645 \times {\frac{\sigma}{\sqrt{n} } } < {\bar X-\mu} < 1.645 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.90

P( \bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } } < \mu < \bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } } ) = 0.90

<u>90% confidence interval for</u> \mu = [ \bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } } , \bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } } ]

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                                      = [$119.86, $123.34]

Therefore, a 90% confidence interval of the true mean is [$119.86, $123.34].

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