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3 years ago

3

Mathematics

1 answer:

snow_tiger [21]3 years ago

7 0

Answer:

12 combinations

Step-by-step explanation:

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The total cost of 5 boxes of pasta is $13. There are 12 ounces of pasta in each box. Each box of pasta cost the same amount. Wha

Bas_tet [7]

Answer: $10.40

Explanation:

First, find the cost of 1 box of pasta. 13/5 = 2.6, so $2.60 a box.

Next, find how many boxes there are. 48/12 = 4, so 4 boxes of pasta.

Finally, multiply 4*2.6 to get the final cost.

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3 years ago

Between which two consecutive integers does ???147 lie?

BartSMP [9]

If those question marks are supposed to be a radical, then 12 and 13. The square root of 147= 12.12, which puts it between 12 and 13.

3 0

3 years ago

Find the median: 7, 4, 5, 2, 7, 7, 2. Need answer asap!!

JulsSmile [24]

Answer:

The answer is 5

Step-by-step explanation:

First arrange the numbers in order :

2, 2, 4 ,5, 7, 7, 7

Then find the middle number :

The middle number is 5 , therefore the answer is 5.

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3 years ago

Name three other real-world situations where a strategy, "plan of attack, or

Dmitry_Shevchenko [17]

Answer:

Planning a wedding .

A huddle in and practice before a football game.

The passing of a Constitutional Amendment.

8 0

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 = $(Mean - \frac{t\times\sigma}{\sqrt{n}})$

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

Lower limit of CI = 98.11

and,

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

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

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

7 0

3 years ago