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strojnjashka [21]
3 years ago
7

Cameron buys 2.45 pounds of apples and 1.65 pounds of pears. Apples and pears each cost c dollars per pound. If the total cost a

fter using the coupon shown is $4.12, write an equation that can be used to find the value of c.
Mathematics
1 answer:
nordsb [41]3 years ago
5 0

Answer:

4.12=2.45c+1.65c

Step-by-step explanation:

We are given,

Cameron buys 2.45 pounds of apple and 1.65 pounds of pears.

Also, the cost of apples and pears is 'c' dollars per pound.

Thus, the cost of 2.45 pounds of apple is 2.45c dollars and the cost of 1.65 pounds of pear is 1.65c dollars.

Since, the total cost after using a coupon is $4.12.

So, we get the equation representing the situation is,

Total cost = Total cost of apples + Total cost of pears.

i.e. 4.12=2.45c+1.65c

i.e. 4.12=4.1c

i.e. c=\frac{4.12}{4.1}

i.e. c = 1 dollar

Hence, the required equation to find c is 4.12=2.45c+1.65c.

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Answer:

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Step-by-step explanation:

D= square root[(y2-y1)^2 + (x2-x1)^2]

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3 years ago
for the data values below construct a 95 confidence interval if the sample mean is known to be 12898 and the standard deviation
Volgvan

Answer:

A 95% confidence interval for the population mean is [3315.13, 22480.87] .

Step-by-step explanation:

We are given that for quality control purposes, we collect a sample of 200 items and find 24 defective items.

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

                             P.Q.  =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~  t_n_-_1

where, \bar X = sample proportion of defective items = 12,898

             s = sample standard deviation = 7,719

            n = sample size = 5

             \mu = population mean

<em> Here for constructing a 95% confidence interval we have used a One-sample t-test statistics as we don't know about population standard deviation. </em>

<u>So, 95% confidence interval for the population mean, </u>\mu<u> is ; </u>

P(-2.776 < t_4 < 2.776) = 0.95  {As the critical value of t at 4 degrees of

                                               freedom are -2.776 & 2.776 with P = 2.5%}  

P(-2.776 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 2.776) = 0.95

P( -2.776 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu} < 2.776 \times {\frac{s}{\sqrt{n} } } ) = 0.95

P( \bar X-2.776 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+2.776 \times {\frac{s}{\sqrt{n} } } ) = 0.95

<u>95% confidence interval for</u> \mu = [ \bar X-2.776 \times {\frac{s}{\sqrt{n} } } , \bar X+2.776 \times {\frac{s}{\sqrt{n} } } ]

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                               = [3315.13, 22480.87]

Therefore, a 95% confidence interval for the population mean is [3315.13, 22480.87] .

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