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mario62 [17]
3 years ago
8

What is the best classification of –60 in the real number system?

Mathematics
1 answer:
Sphinxa [80]3 years ago
4 0
-60 would my classified as an interger
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Jide is n years,his twin sister are two years younger than he is. the sum of their age in years is?
Phoenix [80]
The answer would be n+n+2/2n+2 as Jide is n and his twin sister is two years older so she would be his age (n) plus 2 so n+n+2=2n+2
5 0
3 years ago
80,000,000 in expanded form, word form ,standard form
vampirchik [111]
80,000,000 = 80,000,000 (in expanded form)
80,000,000 = eighty million (in word form)
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3 0
3 years ago
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A flying cannonball’s height is described by formula y=−16t^2+300t. Find the highest point of its trajectory. In how many second
Yakvenalex [24]
<span>Highest point = 1406.25 Number of seconds = 9.375 We've been given the quadratic equation y = -16t^2 + 300t which describes a parabola. Since a parabola is a symmetric curve, the highest value will have a t value midway between its roots. So using the quadratic formula with A = -16, B = 300, C = 0. We get the roots of t = 0, and t = 18.75. The midpoint will be (0 + 18.75)/2 = 9.375 So let's calculate the height at t = 9.375. y = -16t^2 + 300t y = -16(9.375)^2 + 300(9.375) y = -16(87.890625) + 300(9.375) y = -1406.25 + 2812.5 y = 1406.25 So the highest point will be 1406.25 after 9.375 seconds. Let's verify that. I'll use the value of (9.375 + e) for the time and substitute that into the height equation and see what I get.' y = -16t^2 + 300t y = -16(9.375 + e)^2 + 300(9.375 + e) y = -16(87.890625 + 18.75e + e^2) + 300(9.375 + e) y = -1406.25 - 300e - 16e^2 + 2812.5 + 300e y = 1406.25 - 16e^2 Notice that the only term with e is -16e^2. Any non-zero value for e will cause that term to be negative and reduce the total value of the equation. Therefore any time value other than 9.375 will result in a lower height of the cannon ball. So 9.375 is the correct time and 1406.25 is the correct height.</span>
4 0
3 years ago
The accompanying data represent the daily​ (for example, Monday to​ Tuesday) movement of Johnson​ &amp; Johnson​ (JNJ) stock for
egoroff_w [7]

Supposing that the stock increases in 37 days, the 95% confidence interval for the proportion of days JMJ stock increases is: (0.484, 0.7292)

  • The lower bound is of 0.484.
  • The upper bound is of 0.7292.
  • The interpretation is that we are <u>95% sure that the true proportion</u> of all days in which the JMJ stock increases <u>is between 0.484 and 0.7292.</u>

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of \alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the z-score that has a p-value of \frac{1+\alpha}{2}.

Supposing that it increases on 37 out of 61 days:

n = 61, \pi = \frac{37}{61} = 0.6066

95% confidence level

So \alpha = 0.95, z is the value of Z that has a p-value of \frac{1+0.95}{2} = 0.975, so z = 1.96.  

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6066 - 1.96\sqrt{\frac{0.6066(0.3934)}{61}} = 0.484

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.6066 + 1.96\sqrt{\frac{0.6066(0.3934)}{61}} = 0.7292

The ​95% confidence interval for the proportion of days JMJ stock increases is (0.484, 0.7292), in which 0.484 is the lower bound and 0.7292 is the upper bound.

The interpretation is that we are <u>95% sure that the true proportion</u> of all days in which the JMJ stock increases <u>is between 0.484 and 0.7292.</u>

A similar problem is given at brainly.com/question/16807970

4 0
2 years ago
Parisian sewer #1 contains 9 fewer gigantic rats—rats as big as cats, some might say—than parisian sewer #2. parisian sewer #2 c
Leya [2.2K]

There are different kinds of math problem. There will be 11 rats in sewer #1.

<h3>What are word problem?</h3>

The term  word problems is known to be problems that are associated with a story, math, etc. They are known to often vary in terms of technicality.

Lets take

sewer #1 = a

sewer #2 = b

sewer #3 = c

Note that   A=B-9

So then you would have:

A=B-9

B=C- 5

A+B+C=56

Then you have to do a substitution so as to find C:

(B- 9) + (C-5) + C = 56

{ (C- 5)-9} + (C-5) + C = 56

3C - 19 = 56

3C = 75

B = C- 5

B = 25 - 5

Therefore, B = 20

A = B - 9

= 25 - 9

=11

Therefore, there are are 11 rats in sewer #1

Learn more about  Word Problems from

brainly.com/question/21405634

8 0
2 years ago
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