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Tpy6a [65]
3 years ago
12

Isaiah and his children went into a grocery store and where they sell apples for $0.50 each and peaches for $1.50 each. Isaiah h

as $15 to spend and must buy no less than 14 apples and peaches altogether. If Isaiah decided to buy 8 apples, determine the maximum number of peaches that he could buy. If there are no possible solutions, submit an empty answer.
Mathematics
2 answers:
pychu [463]3 years ago
8 0

Answer:

8x0.50=4

15-4=9

9/1.50= 6

6 peaches

erica [24]3 years ago
6 0

Answer:

Maximum of <u>6 peaches</u>, and 8 apples

Explanation:

Given that the quantity of apples and peaches must total to 14, a total cost of $15, with a cost of $0.50 each for apples and a cost of $1.50 each for peaches, we can create a system of equations to solve for each variable.

0.5a + 1.50p = 15.

a + p = 14.

There are many methods that can be used but the easiest one that works for this problem is elimination.

Start by multiplying the top equation by 2 to get rid of the decimals.

0.5 a × 2 + 1.5 p × 2 = 15 × 2.

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Step-by-step explanation:

3 ( 6x + 4 )

3 x 6x = 18x

3 x 4 = 12

18x + 12

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3 years ago
Mike put sports books on 1/3 of his bookshelf. He put history books on 1/4 of the bookshelf . What fraction of the bookshelf did
Shkiper50 [21]

Answer:

66.67% or 2/3

Step-by-step explanation:

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3 years ago
A data set includes 103 body temperatures of healthy adult humans having a mean of 98.3degreesF and a standard deviation of 0.73
faust18 [17]

Answer:

CI = (98.11 , 98.49)

The value of 98.6°F suggests that this is significantly higher

Step-by-step explanation:

Data provided in the question:

sample size, n = 103

Mean temperature, μ = 98.3

°

Standard deviation, σ = 0.73

Degrees of freedom, df = n - 1 = 102

Now,

For Confidence level of 99%, and df = 102, the t-value = 2.62      [from the standard t table]

Therefore,

CI = (Mean - \frac{t\times\sigma}{\sqrt{n}},Mean + \frac{t\times\sigma}{\sqrt{n}})

Thus,

Lower limit of CI =  (Mean - \frac{t\times\sigma}{\sqrt{n}})

or

Lower limit of CI =  (98.3 - \frac{2.62\times0.73}{\sqrt{103}})

or

Lower limit of CI = 98.11

and,

Upper limit of CI =  (Mean + \frac{t\times\sigma}{\sqrt{n}})

or

Upper limit of CI =  (98.3 + \frac{2.62\times0.73}{\sqrt{103}})

or

Upper limit of CI = 98.49

Hence,

CI = (98.11 , 98.49)

The value of 98.6°F suggests that this is significantly higher and  the mean temperature could very possibly be 98.6°F

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

5 0
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