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atroni [7]
3 years ago
8

What two nubers multiply to -10 but add to -3

Mathematics
2 answers:
emmasim [6.3K]3 years ago
7 0
-5 * 2 = -10
-5 + 2 = -3
Hope this helped :)
solong [7]3 years ago
7 0
The answer is -5 and 2 because
-5x2= -10
-5+2=-3
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The diameters of bolts produced in a machine shop are normally distributed with a mean of 5.7 millimeters and a standard deviati
slava [35]

Answer:

The diameter that separates the top 3% is of 5.85 millimeters, and the one which separates the bottom 3% is of 5.55 millimeters.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Mean of 5.7 millimeters and a standard deviation of 0.08 millimeters.

This means that \mu = 5.7, \sigma = 0.08

Top 3%

The 100 - 3 = 97th percentile, which is X when Z has a p-value of 0.97, so X when Z = 1.88.

Z = \frac{X - \mu}{\sigma}

1.88 = \frac{X - 5.7}{0.08}

X - 5.7 = 1.88*0.08

X = 5.85

Bottom 3%

The 3rd percentile, which is X when Z has a p-value of 0.03, so X when Z = -1.88.

Z = \frac{X - \mu}{\sigma}

-1.88 = \frac{X - 5.7}{0.08}

X - 5.7 = -1.88*0.08

X = 5.55

The diameter that separates the top 3% is of 5.85 millimeters, and the one which separates the bottom 3% is of 5.55 millimeters.

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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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3 years ago
A rocket is launched from atop a 55-foot cliff with an initial velocity of 138 ft/s. a. Substitute the values into the vertical
Nonamiya [84]
It would take 9 seconds.

Substituting the values into the equation, we have:
 
0 = -16t² + 138t + 55

The quadratic formula is:
t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

Inserting our information we have:
t=\frac{-138\pm \sqrt{138^2-4(-16)(55)}}{2(-16)}
\\
\\=\frac{-138\pm \sqrt{19044--3520}}{-32}
\\
\\=\frac{-138\pm \sqrt{19044+3520}}{-32}
\\
\\=\frac{-138\pm \sqrt{22564}}{-32}
\\
\\=\frac{-138\pm 150}{-32}=\frac{-138+150}{-32}\text{ or }  \frac{-138-150}{-32}
\\
\\=\frac{12}{-32}\text { or } \frac{288}{32}=-0.375 \text{ or } 9
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4 years ago
Prove that sec^2x-cosec^2x=tan^2x-cot^2x
Agata [3.3K]
Use the identities sec^2(x) = tan^2(x) + 1<span> and </span>cosec^2(x) = 1 + cot^2 (x)<span>, both are derived from the pythagorean identity of </span>1 = sin^2 (x) + cos^2 (x)<span> by dividing through by either sin</span>2<span> or cos</span>2<span> .</span>
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4 years ago
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