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dolphi86 [110]
2 years ago
7

You have $18 to spend on oranges and grapes. Graph the equation 1.5x+2y=181.5x+2y=18, where xx is the number of pounds of orange

s and yy is the number of pounds of grapes. Then find the maximum number of pounds of grapes you can buy.

Mathematics
2 answers:
aalyn [17]2 years ago
8 0

Answer:

Maximum number of pounds of grapes: 9

Step-by-step explanation:

If you spend all the money, $18, in grapes then you will maximize the number of pounds of grapes you can buy. That means, to solve the equation for y, the number of pounds of grapes, with x = 0 (the number of pounds of oranges):

1.5*0 + 2*y = 18

y = 18/2

y = 9

julia-pushkina [17]2 years ago
7 0
The answer is nine pounds of grapes.
X=0
Y=9

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sophie makes homemade pizza dough.The recipe calls for 2 cups of flour for every 1/3 cup of water.What is the rate in cups of fl
Zarrin [17]

Answer:

the rate is: 6 cups of flour per cup of water

Step-by-step explanation:

Recall that rate involve quotient of the two quantities in question:

Then cups of flour per cups of water is the quotient: 2 cups of flour divided by 1/3 cup of water:

\frac{2}{\frac{1}{3} } =2\,*\,\frac{3}{1} = 6

this means 6 cups of flour per cup of water

6 0
2 years ago
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Kayla has 24 yellow beads and 36 green beads a. what is the greatest number of necklaces she could make? b. how many yellow bead
creativ13 [48]
She can make 6 necklaces and these will be 4 yellow and 6 green on each necklace
7 0
2 years ago
A florist gathered data about the weekly number of flower deliveries he made to homes and to businesses for several weeks. He us
shutvik [7]

Answer:

The number of deliveries that are predicted to be made to homes during a week with 50 deliveries to business is 87 deliveries

Step-by-step explanation:

The data categorization are;

The number of home deliveries = x

The number of delivery to businesses = y

The line of best fit is y = 0.555·x + 1.629

The number of deliveries that would be made to homes when 50 deliveries are made to businesses is found as follows;

We substitute y = 50 in the line of best fit to get;

50 = 0.555·x + 1.629 =

50 - 1.629 = 0.555·x

0.555·x = 48.371

x = 48.371/0.555= 87.155

Therefore, since we are dealing with deliveries, we approximate to the nearest whole number delivery which is 87 deliveries.

4 0
2 years ago
HELP ME
Vladimir79 [104]
Alrighty, so, if I remember correctly: For the first question you have 20 possible outcomes, 4 of which are multiples of 5. (5,10,15,20). This gives you 4/20, I multiplied both the numerator and denominator by 200 which then gave me 800/4000. Next I divided which gave me 0.2.
6 0
2 years ago
A random sample of 250 students at a university finds that these students take a mean of 15.2 credit hours per quarter with a st
Orlov [11]

Answer:

95% confidence interval for the mean credit hours taken by a student each quarter is [14.915 hours , 15.485 hours].

Step-by-step explanation:

We are given that a random sample of 250 students at a university finds that these students take a mean of 15.2 credit hours per quarter with a standard deviation of 2.3 credit hours.

Firstly, the pivotal quantity for 95% 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 credit hours per quarter = 15.2 credit hours

             s = sample standard deviation = 2.3 credit hours

             n = sample of students = 250

             \mu = population mean credit hours per quarter

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

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

P(-1.96 < t_2_4_9 < 1.96) = 0.95  {As the critical value of t at 249 degree of

                                        freedom are -1.96 & 1.96 with P = 2.5%}  

P(-1.96 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 1.96) = 0.95

P( -1.96 \times {\frac{s}{\sqrt{n} } } < {\bar X-\mu} < 1.96 \times {\frac{s}{\sqrt{n} } } ) = 0.95

P( \bar X-1.96 \times {\frac{s}{\sqrt{n} } } < \mu < \bar X+1.96 \times {\frac{s}{\sqrt{n} } } ) = 0.95

<u><em>95% confidence interval for</em></u> \mu = [ \bar X-1.96 \times {\frac{s}{\sqrt{n} } } , \bar X+1.96 \times {\frac{s}{\sqrt{n} } } ]

                  = [ 15.2-1.96 \times {\frac{2.3}{\sqrt{250} } } , 15.2+1.96 \times {\frac{2.3}{\sqrt{250} } } ]

                  = [14.915 hours , 15.485 hours]

Therefore, 95% confidence interval for the mean credit hours taken by a student each quarter is [14.915 hours , 15.485 hours].

<em>The interpretation of the above confidence interval is that we are 95% confident that the true mean credit hours taken by a student each quarter will be between 14.915 credit hours and 15.485 credit hours.</em>

8 0
2 years ago
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