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arsen [322]
3 years ago
15

- If you can buy one jar of crushed ginger

Mathematics
1 answer:
Fudgin [204]3 years ago
6 0

Answer:

C, 3

Step-by-step explanation:

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Which set of numbers can represent the side lengths, in millimeters, of an obtuse triangle?
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The answer is actually A.

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Tina is finding the quotient of. Is she correct? If not, what changes should be made?
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For this case, we must find the quotient of

(15x ^ 3 + 12x ^ 2-9x) divided by3x, to verify if Tina was correct.

So:

\frac {15x ^ 3 + 12x ^ 2-9x} {3x} =\\\frac {15x ^ 3} {3x} + \frac {12x ^ 2} {3x} - \frac {9x} {3x} =\\5x ^ 2 + 4x-3

So, Tina was wrong, she had to divide by 3x.

Answer:

Option B


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3 years ago
Complete the equation of variation where y varies inversely as x and y = 80 when x = 0.7.
andre [41]
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Any luck with this question ??
Dennis_Churaev [7]
Linear, decrease by 2 for every decrease of x.
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A certain car model has a mean gas mileage of 34 miles per gallon (mpg) with a standard deviation A pizza delivery company buys
zubka84 [21]

Answer:

z =\frac{33.3- 34}{\frac{5}{\sqrt{54}}}= -1.028

z =\frac{34.3- 34}{\frac{5}{\sqrt{54}}}= 0.441

An we can use the normal standard table and the following difference and we got this result:

P(-1.028

Step-by-step explanation:

Assuming this statement to complete the problem "with a standard deviation 5 mpg"

We have the following info given:

\mu = 34 represent the mean

\sigma= 5 represent the deviation

We have a sample size of n = 54 and we want to find this probability:

P(33.3 < \bar X< 34.3)

And for this case since the sample size is large enough >30 we can apply the central limit theorem and then we can use this distribution:

\bar X \sim N(\mu , \frac{\sigma}{\sqrt{n}})

And we can use the z score formula given by:

z=\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And replacing we got:

z =\frac{33.3- 34}{\frac{5}{\sqrt{54}}}= -1.028

z =\frac{34.3- 34}{\frac{5}{\sqrt{54}}}= 0.441

An we can use the normal standard table and the following difference and we got this result:

P(-1.028

7 0
4 years ago
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